Shape your Career With World’s Highest Paying Jobs for Freshers and Experienced

Choosing a high paying career path requires focus, dedication, and guidance. Schooling is all fine and dandy, but it can’t ensure you’ll make an impact out in the real world. However, the right professional field will push you to learn new skills and help you grow as an individual. While choosing your high-paying career, make sure you make the right decision based on your interests, strengths, passions, and goals. For most people, choosing a job is all about salary, benefits, work-life balance, and culture.

If there’s one thing that’s certain, it’s that everyone wants to make as much money as possible. That means getting a well-paying job that allows you to support yourself and your family whether you want an around the clock job or one that lets you spend time with friends and family.

If you’re looking to make money in the long term, consider one of these highest paying jobs in the world that are in demand. With the world becoming increasingly competitive, your future is in your hands. With hard work and perseverance, you can achieve anything you wish to achieve and we’re here to help. Today, we’re going to cover some of the Highest Paying Jobs in the World in different industries and explained their great benefits.

List of Highest Paid Jobs in the World that would Shape your Future

Job Title Qualification Salary 
Corporate Lawyers
  • B.A. LLB, BBA LLB, from a recognized university
Rs. 20- 50 LPA
Investment Banker
  • 10+2 (preferably Commerce)
  • B.Com/B.B.A Or Bachelors in Economics or Finance
  • Masters in Business Administration (MBA)
Rs. 10-40 LPA
Data Scientist
  • Bachelor’s degree Computer Science/ IT or related field.
  • Master’s degree in Data Science or related field
Rs. 20-50 LPA
Medical Professionals
  • 10+2 with 60% marks in the Science stream
  • M.B.B.S, followed by M.D. or M.S
Rs. 20 – 40 LPA
IT Systems Manager
  • BSIT or B.Sc IT,
  • B.Sc CS
  • M.Sc in Computer Science
  • Advanced Certificate in Cyber Security
Rs. 20 -30 LPA

Now that we’ve glanced at the highest paying jobs in the world in future, let’s examine a few of them in detail.

1. Corporate Lawyers

Corporate lawyers represent companies and businesses on their legal duties and rights. They protect the legality of transactions and represent corps in commercial matters such as corporate tax, mergers, acquisitions, etc.

Communication skills, analytical skills, and business knowledge are all necessary if you want to be a successful corporate lawyer. A corporate law firm or another organization may be the place for you — and your salary will vary accordingly.

When you first step foot into Corporate Law, you’ll often begin as an associate. Associates work under senior attorneys, conducting research and writing briefs. Despite its reputation for long hours and difficult clients, the corporate law profession is one of the most lucrative ones out there.

Big-name firms offer generous salaries, stock options, and bonuses to new associates right out of law school. Though the challenges of the job are many, at the end of the day it’s all worth it.

Education

  • To become a corporate lawyer, future candidates should study law for three years as a graduate.
  • Students can also opt for an integrated course of five years, where they earn a Bachelor’s degree alongside their law credentials.
  • After 10+2, you can also register for CLAT (Common Law Entrance Test) and appear in the entrance examination offered by law colleges and universities. Admission is given on the basis of the score that you get in the entrance test conducted by law schools.

Career Prospects 

After globalization, corporate law has become one of the most sought-after career options.

  • Corporate lawyers are in high demand across all fields — whether it is an organization’s corporate governance or its business consulting services.
  • The fast pace of globalization has made foreign law firms move to India. This has, in turn, increased the scope of opportunities for Indian corporate lawyers.

Salary

Corporate lawyers in Government enterprises can expect a salary of Rs.20 lakh per year on average. However, the financial success of a corporate lawyer in the private sector is based on a number of factors, including his experience and knowledge of the field.

Top recruiters

  • Khaitan and Company,
  • Amarchand Mangaldas and Co
  • DSK Legal
  • McAfee
  • Cisco
  • IBM
  • Boeing
  • Intel

2. Investment Banker

The banking industry in India has a growing number of public, private, and foreign-run financial institutions that rival the industries of most major economies. There are 27 public-sector banks, 21 private-sector banks, and 49 foreign-run banks in the country.

In addition to these, there are many regional rural banks and thousands of urban and rural cooperative banks. An investment banker is a professional who specializes in helping companies to go public, selling shares to individual investors.

This highly sought-after job is the key to advancing the careers of young professionals because it affords them stable, lucrative careers in finance. In the recently concluded fiscal year, public sector banks in India accounted for a staggering 61.21 percent of total banking assets in the country.

Asset diversification is a rather new phenomenon in the Indian banking industry. Investments have been steadily growing from being a negligible part of bank balance sheets just a few years ago to becoming the second-largest constituent today.

Education

  • People in the profession of financial advising often hold degrees in economics or finance;
  • The Chartered Financial Analyst (CFA) designation is designed for people who want to break into financial advising.
  • An accredited bachelor’s degree in finance may also be necessary.

Career Prospects

  • An investment banker starts their career on the analyst track, gaining work experience on all levels of a deal from an idea to a final close.
  • Associates will typically work on multiple deals at a time. The associate track is a step up from analyst, and associates can move into more senior roles such as vice president, supervising analyst and associate banker.
  • The Managing Director (MD) is the highest level an investment banker can attain. At this level, investment bankers are responsible for introducing new ideas that will birth new businesses for the bank.

As investment bankers progress further in their careers, they are rewarded with higher salaries, better benefits, and titles that reflect their seniority — because the firm wants to keep them around.

Salary

Investment bankers in India make Rs. 942,390 annually on average, with the highest earners bringing home over Rs. 1.7 crore per year. In the US, investment bankers make an average of $101,465 a year with the highest-paid earning over $165,000 a year.

Top Recruiters

  • HDFC
  • ICICI
  • Barclays
  • Morgan Stanley
  • HSBC

Do Check: Top 10 Highest Paying Jobs in India

3. Data Scientist

Data scientists are data experts who analyze data by using their knowledge of statistics, programming, and mathematics. They help businesses to use available information to find the most feasible solution.

Data scientists are the go-to people for businesses that need help making sense of data. They are in-demand professionals who are sought after by major players across industries, including retail, healthcare, media & entertainment, transportation, education, and BFSI.

The primary responsibility of a data scientist is to scan a vast amount of data and analyze the gathered information to help a company reach its goals. A data scientist may use many different methods to gather information, from examining the public’s opinion on social media profiles to studying customers’ behavior in-store or online.

The insights gained from this research lead to actionable strategies for increased sales and productivity.

Education

  • Data scientist requires an in-depth knowledge of coding, statistical analysis, and machine learning.
  • A master’s degree in computer science or other quantitative degree is required for this high-paying job.
  • After a Master’s degree, one can further their education with an Executive Post-Graduate Program in Data Science.

Career Prospects

  • Data science is evolving as a field. In the past, it was considered as a job of its own, but today it’s more about an approach towards technology.
  • Industries have started realizing the importance of analytics and data, and have started focusing on engaging more data scientists to work with them.

Salary

If you’re passionate about data, math, and science, then the role of a data scientist is for you. Data scientists are in huge demand globally, with an average annual salary of Rs.822,895 in India and $96,501 in the US.

Top Recruiters

  • Couture
  • Mate Labs
  • Amazon
  • Sigmoid
  • IBM

4. Medical Professionals

Medical professionals are the first-line healers in our society. Whether they are called doctors, surgeons, nurses, or medical lab technicians, their job is to help people get better when they are sick. They are tasked with diagnosing illnesses and prescribing effective treatments to ease the patient’s discomfort.

Becoming a highly skilled medical professional takes an incredible amount of effort and years of study. To become a well-rounded doctor in the field, one must complete many in-depth internships, which are in turn rewarded with high pay. It’s no surprise that the highest paying jobs in the world are all medical professions.

The job of a medical professional is to care for patients with empathy and knowledge. They must examine patients thoroughly to discern the cause of their ailments. Armed with this information, they can prescribe medication and act as guides through complicated procedures.

Education

  • 10+2 with Science PCB/ PCMB
  • M.B.B.S, followed by M. D or M. S

Career Prospects

The field of medical professionals offers endless opportunities for professionals looking to make a difference in the world. Those with bachelor’s, master’s, PhDs, and MDs can find satisfying careers in both the private and non-profit sectors. Opportunities cover everything from business management to education, whether it’s working for a hospital or a non-governmental organization.

Salary

The average annual pay of a medical professional in India is 1.1 million rupees, or about $15,000 USD. Meanwhile, general surgeons in the US make about $300,000 per year or an approximate monthly payment of $2,500.

Top Recruiters

  • AIIMS (All India Institute of Medical Sciences)
  • Apollo
  • Fortis
  • Max
  • Columbia Asia

5. IT Systems Manager

The Indian IT industry is among the fastest-growing sectors in the world. India has the highest share in the global services sourcing industry, is the second-largest hub for tech startups and has the second-largest number of internet subscribers in the world.

India offers several advantages that make it an attractive outsourcing destination, including its access to a large (and growing) pool of skilled labor, broadband connectivity, and relatively low labor costs. Businesses in this sector include providers of cloud services, technology solutions, and business process management.

The IT sector has a large pool of available talent, providing rich ground for IT professionals. And the scope of opportunities is plenty:

  • One can work in developing small-scale software solutions
  • Manage large-scale infrastructure projects,
  • Run an IT consultancy firm, or
  • Create new software for a service provider

Education

  • Must have a graduate degree in computer science or information technology .
  • Courses like BSIT or B.Sc IT, or B.Sc CS will allow you to pursue this career path, ensuring you’ll have the skills and real-world experience you need for advancement.
  • Apart from a bachelor’s degree, internships can further boost your career prospects, as they let you gain hands-on experience after college graduation.
  • A master’s degree such as M.Sc in Computer Science and an Advanced Certificate in Cyber Security will help you even more

These advanced certifications show will help you learn and improve, which is an excellent quality that most employers look for in new employees.

Career Prospects

  • IT systems managers are in charge of the company’s computer-related tasks, including software installation, analysis, and support. They are also tasked with overseeing information security.
  • They are often required to manage the IT department, which involves strategic planning for operating budgets, working with vendors on purchases, and making curriculum decisions.
  • In a larger organization you could be promoted to work on some of the world’s most challenging projects, manage complex IT projects, or even manage other members of your team. With enough exposure, you could even become a leading light in IT management.

Salary

While the average pay of an IT systems manager is Rs.809,969 in India, it is US$87,437 in the United States.

Top Recruiters 

  • TCS
  • Cognizant
  • Infosys
  • Dell
  • SAP
  • IBM

Conclusion

While these are some of the highest paying jobs in the world, numerous other high paying careers have excellent job security and are just as lucrative. The key, however, is to choose a career that not only pays well but also allows you to grow professionally. Career satisfaction is a combination of a wide range of factors, but passion and growth potential are chief among them.

When you’re finding a career, look for one that will allow you to grow in both your skills and your paycheck. If you’re passionate about something, there’s no better feeling than being able to turn that interest into a part-time or even full-time job.

Best Online Computer Courses to Get a Job | Top 10 Computer Courses in Demand in India

Best Online Computer Courses to Get a Job

Best Online Computer Courses To Get a Job: The era of technology gives a massive wave of opportunities to achieve high academic qualifications for millennials and generations Z. For most engineering students, computer courses are one of the most popular alternatives. Well, it is possible to separate computer courses into several areas. Your career will be determined and the correct method chosen. Every year demand for online computer courses increases as the numeration spreads across the country.

Computer students have perfect pay that draws young students to pursue their careers in computer science. If you’re hard on deciding what program or striving to acquire high promotion requirements, you’re going to have the top online computer courses in 2020 to promote an excellent future career.

Top Online Computer Courses to get a High Paying Job

There are a variety of online computer courses available before being added to the list. Choosing the correct computer course to enhance your profession is vital to determine your strength, weakness, and abilities. In addition, online computer courses are an advantage for students when graduates are not sufficient in a competitive economy.

Cybersecurity

Cybersecurity is necessary for everyone who carries out their financial transactions online. According to Statista, the highest demand for occupations in the year 2020 is cybersecurity. You can take a cybersecurity course if you’re interested in the Internet. Every transaction is made online or via its mobile app, from the finished technology to the eCommerce market. To prevent theft, Internet attacks, and unsuspected ransomware, it is essential to create a high level of cyber protection.

You can run a cybersecurity course online to learn the nature of cyber assaults, identify hazards online and take preventive action. In this course, you will learn how to evaluate technical risks and take preventative measures to avoid various online threats.

Data Science

Based on the LinkedIn posting, India is predicted to have around 150,000 employment by the end of 2021 with a data scientist, which amounts to over 62 percent in 2019. Data scientists search for their analytics in industrial areas such as drugs, aeronautics, oil, energy, and marketing. With only three years of experience, data scientists can earn up to Rs. 20.000 per year, making data science one among the most efficient online computer classes for high-wage work.

The course includes mathematical, statistical, photonic, advanced statistics in Python, deep learning, machine learning, and data science education. Thus, the course gives you a whole toolkit to be a skilled data scientist. You also receive instruction in debunking and typical misunderstandings from the data pool.

Big Data Engineering

As its name implies, Big Data is a big pool of data methodically analyzed and used to collect data. Today, when consumers choose to transact online, big data plays a critical role in helping organizations understand how consumers spend online.

You have to design, maintain, assess, and test Big Data solutions to follow the extensive data engineering program. The primary task for a prominent data engineer is to build large-scale data processing systems. If you also have an expert in data analysis and storage systems, you will be able to work with the latest data technologies such as R, MySQL, Python, NoS QL, etc. The leading Data Engineer can earn up to $146,000 a year, unusually the highest average wage for IT workers.

Do refer: Top 10 Highest Paying Jobs in India

Data Analyst

To extract the data from the unstructured data pool, a data analyst must apply correct statistical and logical thinking methods. You need to help organizations do qualitative research to provide structured data, which they can then employ to conduct their commercialization, administrative, and production activities. This is very closely related to data science, although the job of the data analyst and the job of the data scientist is very different.

In contrast to data scientists, considerable coding and programming are not necessary. However, you should know the latest data technologies and languages such as Python, Ruby on Rails, R, MySQL, etc. The data analyst can make up to $105,000 per year and up to a minimum of $69,000, according to Glassdoor.

Big Data Analysis

Data analysts can comprehend the newest trends in the industry and relevant views of the large pools of data. Many companies, including Fintech, can recruit you based on relevant big data to make informed judgments and improve company procedures. In addition, you can obtain important online and offline information from various activities, such as email marketing, digital marketing, social media marketing, etc.

You need to grasp several techniques of storing data in Big Data Analysis. The plan also includes data from unstructured to structured data processing and analysis. This program prepares you for all your skills to analyze, process, manage and store unstructured and structured data for corporate purposes. According to the fact, the data analyst might generate up to $139,000 for his functions and obligations annually.

Web Designing

Suppose the web design course is your favorite one if you enjoy visuals. Web design encompasses the development of stunning designs, including logos, web designs, Websites, brochures, and other graphics and printed media elements. In addition to the visuals, you must master HTML, XHTML, Javascript, CSS, and more programming languages.

With increasing experience with mobile devices, you also need to create websites to improve the user experience and interfaces. The website design program covers all websites, including web design, HTML, PSD to HTML, and PHP. It would allow you to finish your design with programs such as Adobe Photoshop, Corel Draw. A web designer’s average annual pay is about $64,000.

VFX Training And Character Animation Degree

The technology that focuses on smart devices and gadgets delivers a new era of games and films. The gaming industry is driven by increased reality and virtual reality, giving rise to modern technologies, particularly VFX and animation of characters.

You will also study advanced VFX effects, software skills, and technology extensively used in making movies during the VFX training course. A VFX artist’s scope is enormous because films, animated films, and games are increasingly demanded. Also, like in the game “Pokemon Go,” you can test character animators to create unique characters. An artist with 5 to 7 years’ experience has an average annual wage of $75 000, and the character animator makes $65,000 annually.

Software Development

Software development demands are rising, and mobile apps and online apps are emerging. A developer works with C++, Java,.NET, ReactJS, and many different programming languages. In addition, you must produce mobile applications, online applications, and software that is valuable for companies as a software developer.

You must focus on the current coding methods and development methods such as Hadoop, JavaScript, iOS, Android, Unity, and many other software developments and programming languages courses. Developers of software can make up to $108,000 per year.

Computer Hardware Engineering And Networking

The main focus of computer hardware and networking is on the maintenance and repair of critical computer elements. The constant use of equipment causes you to work with prevalent challenges. Predictive and preventive measurements can be strategized to avoid significant equipment maintenance.

You will learn several topics such as LAN, DTP, CCNA, and many others in computer hardware and networking courses. A computer hardware engineer’s average income is about $114,000.

Conclusion

These online computer courses for your work are available to you, which save your time and transit costs and make your routine flexible. You must now have a reasonably solid understanding of the type of online computer training you can take. The demand for these courses in India is substantial.

The ultimate decision of online computer courses depends on your profession and the type of jobs you want. Once you understand both, it will be much easier for you to choose the course.

Top 10 Highest Paying Jobs in India That You Should Consider | Complete Guide on Highest Paying Professions in India

Top 10 Highest Paying Jobs in India

In search of the best jobs in India? The ones that pay well and offer you a great career? If yes, then check out our list of the Top 10 Highest Paying Jobs in India. Luckily, many jobs in India have high payouts across many sectors, from banking to marketing to technical support. Money isn’t the only reason to choose a certain career path, but it can help you make an educated decision about your future.

The job market is one of the most competitive fields for the average Indian student. The only thing that makes students feel good about their efforts is a high-paying career that rewards their hard work with enough money to live off. If money is an important factor for you when choosing a profession, then you’ll be glad to know that these are the highest salary jobs in India.

This list is updated every year, and it has the Best Jobs in India for anyone who wants a different and satisfying career. It can help you decide what job is right for you — if you’re just starting your career or looking for a new challenge. Take a look at each job and use it to open doors to the best-paying jobs in India.

Best Jobs in India

The jobs that once were reserved for people with years of experience are now open to freshers with the right mix of skills. It’s becoming increasingly possible to find jobs that allow you to be creative, work remotely, and earn enough to cover your bills. Consider working in one of these creative fields if you’re looking for a creative career that doesn’t break the bank.

Thanks to the changing landscape of the global economic environment, there has never been a better time for individuals who are passionate about their industry to get involved and pursue their dream job. The following information uncovers the best industries for making money across the world.

Top 10 Highest Salary Jobs in India

Pursuing a career in one’s passion is not only rewarding, but it could also be financially lucrative. With so many options available to students, there is no reason for them to settle for anything less than their dreams. And that’s good news because the job market is more promising than ever — there are more opportunities for young graduates than ever before.

So, here are the top 10 highest paying jobs in India that students should consider the following:

  1. Cloud Architect
  2. Chartered Accountant (CA)
  3. Healthcare Professionals (Doctor and Surgeons)
  4. Investment Banker
  5. Blockchain Developer
  6. Data Scientist
  7. Full Stack Software Developer
  8. Machine Learning Experience
  9. UX/UI Designer
  10. Digital Marketing

1. Cloud Architect

A cloud architect is an emerging field that both cloud software developers and enterprise I.T. departments desperately need it. With ever-changing cloud storage, systems, social media networks, and application deployment environments, cloud architects are able to take the lead in public, private, and hybrid cloud system development.

You can learn more about becoming a cloud architect on Wikipedia or by visiting online courses at sites like Coursera or Edx. Cloud architects hold a crucial place in the I.T. department, building and deploying applications. In their position, they need to be aware of all activity within the cloud, including monitoring cloud activity and keeping track of any changes.

The successful cloud architect should have a proven ability to design and migrate applications while also having a comprehensive knowledge of security mechanisms. Finally, it’s essential that the cloud architect directly reports to senior-level management.

Qualification

To be a successful Cloud Architect, you need to understand the inner workings of application development and the business side of I.T. strategies.

  • A candidate should have skills in networking and managing databases, and he or she must be able to communicate with developers and other I.T. staff.
  • You need experience with monitoring, troubleshooting, and repairing cloud servers to be a good fit for this job.
  • It’s also helpful if you’ve had experience writing cloud-native applications because you’ll be building microservices.

With its ability to brew tea, mix tea, and garnish your favorite drink, it can take away all of your stress with just a touch of a button.

Salary

Cloud architects in India earn a base salary of Rs 1.6 to 2.4 million a year. With experience, this average can increase. The industry standard for experienced cloud architects is a base salary of Rs 1.8 to 3 million a year.

Top companies hiring for this role:

  • IBM
  • Microsoft Corporation
  • Hewlett-Packard
  • EMC
  • Amazon

Do Refer: Best Online Computer Courses to get a Job

2. Chartered Accountant (CA)

Chartered Accountant is a demanding career that is essential for any business to thrive and is one of the highest paying jobs in India. With the complex laws and regulations constantly changing, Chartered Accountants are more important than ever. At the same time, it’s not an easy job to excel in — with little room for error, being a C.A. can be challenging.

Chartered Accountants are members of the Institute of Chartered Accountants of India (I.C.A.I.) — the only national accounting body in the country. With over 1.25 lakh active practitioners, I.C.A.I. is responsible for managing more than 2.82 lakh Chartered Accountants — making it one of the largest accounting bodies in the world by membership numbers.

Qualification

  • Commerce graduates and postgraduates having a minimum of 60% marks and intermediate level passed students of Institute of Company Secretaries of India and Institute of Cost Accountants of India can pursue the Advanced course offered by I.C.A.I.
  • Grad students and postgraduate students can still apply for the C.A. course, though they’ll need a minimum 60 percentile.

Salary

Starting at an average salary of Rs. 6-7 L.P.A., Chartered Accountants can earn up to Rs. 30 L.P.A. or more, depending on their expertise and experience.

Top Companies

  • Standard Chartered
  • K.P.M.G.
  • Ernst & Young
  • Deloitte
  • B.D.O. International
  • Grant Thornton International

3. Healthcare Professionals (Doctor and Surgeons)

In India, healthcare includes both private and government services. Thanks to the rapid advancements in technology and an increase in the number of people falling ill, there is a rise in demand for quality healthcare services that are also affordable. In India, the healthcare industry consists of a few different components: public and private services, technology, and disease incidence.

With good salaries and growing industry, it’s no wonder why so many people choose to become doctors and nurses. As of today, India is the third-largest healthcare market in the world — but with the rapid expansion of this industry, it will take over China and Japan within three years to become number two.

Domains such as healthcare administration, nursing, and medical assistant have a demand for a wide range of medical professionals. Alongside traditional specializations such as dentistry, optometry, and pharmacy, these areas are some of the most promising and high-paying healthcare avenues.

Qualification 

  • Graduation from a 12th-grade board with at least 60% marks in the science stream is required for this job. This includes physics, chemistry, and biology (PCB), or chemistry, mathematics, and biology (PCMB).
  • To enter a reputed medical institute, you have to pass the relevant entrance exams, such as NEET, AIIMS, JIPMER, and other similar entrance examinations conducted by state governments.
  • In order to practice medicine in the country, all doctors must have an M.B.B.S. from a recognized institution run by the Medical Council of India.
  • Students can choose to study M.D. while pursuing their M.B.B.S. degrees and selecting specializations like surgery (M.S.).

Salary

The average salary for a medical professional in India is around 10 LPA, while more than 25% of medical professionals earn more than 20 LPA. The average salary for a General Physician is Rs. 6,95,239, but a General Surgeon earns an annual package of more than 11 LPA.

Top companies 

  • AIIMS (All India Institute of Medical Sciences)
  • Apollo
  • Fortis
  • Max
  • Columbia Asia

4. Investment Banker

An investment banker is an ideal candidate for anyone who loves profit and has a knack for making investments. While there’s a lot of competition in the industry, you’ll be capitalizing on opportunities and advising companies about their finances. On top of offering great pay, this profession has a high potential for growth in the coming years.

Today, the banking industry in India is one of the strongest in the world. The financial services industry in India is growing at a fast pace with foreign banks showing keen interest in partnering with Indian banks to reap the benefits of one of the fastest-growing financial markets. The financial services sector in India, in particular investment banking, has been growing steadily for years and is one of the best jobs in India.

Increased investments in the country’s economy mean higher capital flows, all of which need to be efficiently managed by investment banks. Total investment banking revenue in India broke past the billion-dollar mark for the first time in the first half of 2019 (H1CY19), reaching $523 million; four times what it was in H1CY18.

Qualification

  • Investment bankers need to have a strong understanding of economics, financial markets, real estate, and business. The perfect candidate should also have a bachelor’s degree in finance, economics, mathematics, or business administration.
  • If they have a master’s degree in any of these specialisations that will help them get hired.

Salary

Investment banking is one of the most lucrative careers in the country today and is one of the best in the list of top highest paying jobs in India. Entry-level workers can expect to earn a salary ranging from Rs. 4 lakhs to Rs. 12 lakhs per annum, while experienced employees might bring home even more — up to Rs. 40 lakhs per annum.

Top companies 

  • Goldman Sachs
  • JP Morgan Chase
  • Deutsche Bank
  • HSBC
  • Citibank
  • Axis Capital Ltd.
  • ICICI Securities Ltd.
  • IDBI Capital

5. Blockchain Developer

‘Blockchain Developer’ is one of the most in-demand positions in the world today. At the same time, it is also one of the highest paying and one of the best jobs in India. Blockchain technology has branched out to integrate with both Machine Learning and Artificial Intelligence; people with knowledge of this technology are highly sought after by companies that want to develop their own blockchain for storing data.

There are 2 million software developers in India, but only 5,000 of them have the skills to build blockchain applications. This gap between demand and supply is projected to increase, given that talent is in short supply. Blockchain technology has attracted investors from around the world who are looking for developers with the expertise to build “the next big thing.” A company is just as, if not more, valuable with a good team behind it.

Qualification

To be eligible for this developer position, you must have a BE/BTech in Computer Science/Mathematics/Statistics/Information Technology. Candidates with prior coding experience in popular languages like Java, JavaScript, C#, C++, and Python are preferred in this company.

Salary

The starting salary ranges between Rs. 3 lakhs to Rs. 6 lakhs per annum, while the average salary currently for blockchain technology experts is around Rs. 8 lakhs, making this one of the highest paying developer jobs in India. Not only do new graduates have an opportunity to enter the sector, but blockchain also rewards years of experience with higher paychecks.

Top companies hiring for this role:

  • Auxesis
  • OpenXcell
  • Elemential
  • Sofocle
  • Primchain
  • Signzy
  • SoluLab
  • MindDeft

6. Data Scientist

Data scientists are in demand everywhere — from big corporations to small startups, and data scientist jobs (especially remote ones) can be quite lucrative. With rapid growth and promising salaries, this profession is a great choice for anyone who’s interested in working with large amounts of data and creating comprehensive analytics.

A data scientist is a new breed of executive that needs to be at the helm of every company today. This role is an amalgamation of multiple skills — statistical analysis, machine learning, and more — and it’s specifically needed to analyze and present large sets of data.

Qualification

To succeed in this role, candidates must have at least an undergraduate degree in Engineering (BE/BTech) or Computer Science. Certification in data science will definitely be helpful. Additionally, candidates with a rudimentary knowledge of basic programming languages such as Python, SQL will have an added advantage.

Salary

A data scientist with five years of experience makes between Rs. 40 lakhs to Rs. 50 lakhs per year. The pay for professionals with less than two years of experience varies, so there’s no standard salary for this position — it can range from Rs. 25 lakhs to Rs. 40 lakhs per annum, making it one of the highest paying IT jobs in India.

Top companies 

  • Amazon
  • Walmart Labs
  • GreyAtom
  • Procter & Gamble

7. Full Stack Software Developer

A full-stack developer is a must for companies and organizations that need to build and maintain an e-commerce site. Working with both the frontend and backend development of an online store, these specialists ensure brand consistency and manage product categorization — as well as security, testing, and SEO — to ensure a high-performance website that lives up to their client’s needs.

Even by conservative estimates, the Indian IT/ITeS industry is expected to bring in revenue of $350 billion by 2025. This is only the beginning — demand for software developers in India will continue to skyrocket in the coming years. The economic impact of this new video conferencing technology on society promises to be profound. Established players are scrambling to get a foothold in the new market before it’s too late.

Qualification

  • To become a Full Stack Developer, you will need an undergraduate degree in computer science or a related field such as information technology or engineering.
  • All Full Stack Developers should be well versed in frontend and backend languages like HTML, CSS, Python, C#, Ruby, JavaScript, and SQL.

Salary

Entry-level developers on the Fullstack team earn ₹375,000 per annum. Developers with 1 to 4 years of experience on the Fullstack team earn ₹553,000 per annum. Mid-level developers on the Fullstack team earn ₹1,375,000 per annum.

Top companies 

  • Dell
  • Siemens
  • E2logy
  • Barclays
  • IBM
  • Simpalm
  • ChromeInfotech

8. Machine Learning Experts

Machine learning is an artificial intelligence technique widely used across industries to make business processes smarter, faster, and more user-friendly. With the advent of chatbots, machine learning has become even more accessible — by understanding what customers want and automatically providing it, brands can increase their consumer base, while concurrently increasing customer satisfaction.

Machine learning and AI adoption in India has grown immensely over the last couple of years, and the future is even more exciting! There’s been a sustained growth in AI and machine learning investments — the market is projected to expand at a Compound Annual Growth Rate (CAGR) of 33.49% between 2019 and 2023. The latest stats suggest that the market will be worth $66 billion by 2023.

Qualification

The key to success in this position is a broad and solid background in mathematics, statistics, and computer science. Any experience with data analysis and data mining projects will also be beneficial.

Salary

If you’re looking to take home high paychecks, consider a career in machine learning. Machine learning specialists take home over ₹ 24 lakh per annum on average in India, according to data from PayScale. With a median salary of INR 700,000 (USD 10,944), Machine Learning Engineers in India make good money.

Top companies

  • Accenture
  • Zycus
  • hCapital Business Consulting
  • Quantiphi
  • IBM
  • ITC Infotech

9. UI/UX Designer

A user interface designer has the unique position of being able to see all of a product from many different perspectives. This is why they have to think about more than just how a screen will look — they have to consider how a new feature or design will affect a user’s overall experience, and if a change would be a step up or a step back for the company.

A product’s marketing strategy heavily relies on the visual presentation of its UI/UX design. That’s why companies are eager to hire UI/UX designers — whether you’re directly responsible for designing products, or you’re responsible for designing experiences, your career will benefit from specialized knowledge in this industry.

Qualification

  • Students with 10+2 degrees can opt for undergraduate programs in UI/ UX, provided they have secured at least 50 percent marks in their board exams.
  • For pursuing postgraduate degrees in UI/ UX, students must hold a Bachelors or Master’s degree in any field of study.

Salary

For UI/UX designers in India, the average starting salary is Rs 20,000 to 25,000 per month with experience ranging from 2 to 4 years. On average, an experienced designer in India will earn around Rs 70 lakh in a financial year.

Top companies hiring for this role:

  • Microsoft
  • Adobe
  • Google
  • Accenture
  • Cognizant Technology Solutions

10. Digital Marketing

For businesses, digital marketing is a necessity — not a luxury. In today’s online-dominated marketplace, companies need to update their marketing strategies to keep up with consumer trends and expectations. Digital marketing is here to stay, and brands are making the most of it.

Digital marketing channels are changing the way of communication today. Through digital marketing, brands are able to get in touch with their customers directly, giving them the ability to create intimate connections. The diversity and creative freedom offered by digital marketing draw people from all backgrounds to pursue careers in the field.

There are multiple roles available for those looking for jobs within the digital marketing sphere, including content writers, SEO analysts, social media managers, brand marketing managers, PPC marketers, and many others.

Qualification

While a digital marketing degree isn’t required, having an MBA or other relevant educational background in the field helps. Also, there are a number of online courses available that can help marketers and entrepreneurs to hone their skills and stay abreast of the latest trends in the industry.

Salary

Entry-level employees—those with less than two years of experience—make between Rs. 2 lakhs to Rs. 3 lakhs per year, while professionals with over five years of experience can expect to draw salaries of Rs. 8 lakhs to Rs. 10 lakhs per year. Additionally, those who have worked for over ten years can enjoy annual earnings of Rs. 15 lakhs to Rs.

Top companies 

  • Google
  • Accenture
  • Amazon
  • Red Ventures
  • Cognizant Technology Solutions

Wrapping up

These are the Top 10 Highest Paying Jobs in India — whatever industry you choose, you can climb your way to a lucrative salary. The list contains representatives from diverse professions, from the insanely creative world of arts and entertainment to the science-driven field of medicine and beyond. The prospects of great pay and a fulfilling job are available in every field. If you want to make money and enjoy your job, you need to decide on your goals, choose the right path to get there and stay focused all the way through.

Top Commerce Project Topics & Ideas for Students | Current Topics Related to Commerce for Project

Top Commerce Project Topics and Ideas

Commerce is one of the key elements behind the success of every business. To execute smooth trading operations and maintain profit, a company must focus on commerce, which deals with many things, not just selling and buying. As a fresher, you need to have a proper understanding of every aspect of commerce. In this article, we have listed out some commerce project topics that will help you learn more about commerce.

With the advancement of technology, the world is moving towards digitalization which is why we are getting to see a lot of e-commerce solutions. The introduction of machine learning and AI has made e-commerce more efficient and reliable. E-commerce has now become an important topic for students pursuing graduation and postgraduation.

Students are now required to work on multiple commerce project topics based on e-commerce to experience trade-related issues and explore the operational challenges. When they resolve these issues practically, they will get to learn a lot of skills.

In this article, we have included multiple commerce project topics & ideas for you so that you can work on a good project.

Top 10 Commerce Project Ideas and Topics

In every business, the end customers are considered as Kings; a business needs to analyze their preferences and needs before formulating any marketing strategies and starting any manufacturing. To make sure that this chain of processes works smoothly, we need to rely on commerce.

If you are grinding your brain to come up with ideas that can help you achieve your goals in the commerce field, then you are at the right place because we are going to list out some of the best commerce project ideas. When you work on these ideas, they will not just get you to experience different problems but also guide you on how to resolve them. On top of that, this will be an addition to your resume, thus increasing your chances.

Impact of Outsourcing Material Availability Decision-Making

Objective:

To identify the criteria that are used when a business plans to outsource material availability and strategize the whole outsourcing process.

outsourcing is considered to be beneficial for every business entity, but it comes with a lot of challenges. Usually, businesses opt for outsourcing when they want to minimize their labor cost. However, a business has to take care of a lot of aspects before getting into outsourcing.

Although outsourcing has different levels of challenges and is a completely complex process, if you strategize properly and formulate an effective pre-plan to control outsourcing, it can prove advantageous to your business. First, you need to understand there is no connection between the company’s performance and the outsourced aspects; they are completely independent of each other because the material availability is not affected directly.

The two main challenges that you might face while opting for outsourcing are:

  • First will be time; this is when the production gets slowed down because of the unavailability of materials.
  • Second, the financial stability of the business; if it’s weak, then that might be challenging.

By working on these commerce project ideas, you can come up with solutions to resolve the above challenges. You will also get practical knowledge on outsourcing that will help you in the future.

Improving Employees Performance By Monetary Incentives

Objective:

To perceive the relationship that exists between employees’ performance and monetary incentives.

You must have noticed that in every industry, the bonuses are fixed, and they do not change considering the hard work of employees. If an employee performs more than expected and achieves the target then he/she is entitled to a fixed bonus. Hard work is not considered in the case of fixed bonuses. Therefore, it is clear that monetary benefits don’t depend on the employee’s performance.

One of the most annoying reasons that make an employee demotivated is when the company delays the release of bonuses to its employees. In such situations, the employees become dissatisfied and lose the urge to work harder and earn better.

In some companies, employers quote the bonus in advance to motivate the employees. In such cases, employees are required to achieve the set goals before a fixed time in order to be eligible to get the quoted bonus. This way, employees can perform more efficiently and proactively.

When you work on these commerce project ideas, you will get to learn a lot of things about monetary benefits, but most importantly, you will learn how incentives can motivate an employee and create an urge in them to perform better.

Outsourcing Human Resource in Beverage and Food Firms

Objective:

To understand the end results when a company outsourced human resource functions on organizational performance. This will include predicting the potential benefits that will come from this decision and also how much effect it will put on the employee’s performance in the long run.

When a company is not able to allocate a budget for a separate human resource department, then it leaves the company with one solution, which is to take the help of outsourcing. This might decrease the company’s liability for its working employees, but this hasn’t proved beneficial in the long run. According to many reports, it is difficult to have a focused career in outsourcing.

The main reason that makes a company opt for outsourcing human resources is cost. Outsourcing might help in reducing the cost of a company and also in bringing some productive resources to the company. However, you won’t be able to enjoy these benefits in the long run.

It is noticed that outsourcing human resources can lead to a disconnection between the employees. This will reflect in their attitude because most of them start working irresponsibly when some task is assigned to them. It is observed that employees have a high nutrition rate during these times.

This commerce project idea is best for food industries because, in these industries, the nature of commerce is completely time-sensitive, which can be a challenge for human resource professionals.

Role of E-commerce in Reducing Operational Cost

Objectives:

To examine and evaluate how e-commerce manages to reduce the functional cost with the help of comparative analysis

When you start working on this commerce project topic, you will get to know the benefits of e-commerce in today’s world. During the pandemic, the rate of electronic transactions has increased exponentially. An organization that operates from a physical office has to bear more expenses compared to those organizations that are working remotely.

Some of the costs that one has to bear when they operate through a physical office are administration, logistics, salesperson salary, electricity, lease or rent, telephone, taxes, heating & cooling, and repair & maintenance of the building. On the other hand, when organizations operate remotely, they only have to bear the cost of shopping cart software, web hosting charges, and distribution rates.

It is now clear that maintaining a physical office is expensive. So when organizations shift their business to e-commerce, they might reduce their functional and operational cost. On top of that, e-commerce has a lot of hidden benefits that will definitely help the organization to grow.

Reducing Unemployment Through a Co-operative Movement

Objective:

To create and establish a cooperative movement in a working society in order to analyze its benefits on society members. It also examines various aspects of the cooperative movement, which is responsible for providing support to team members.

When you start working on this commerce project idea, you will come to know the benefits of a cooperative movement and how it helps to reduce unemployment. According to research, it was concluded that cooperative movements always support employees and their well-being in terms of career growth and education. This is why every organization should think of adopting it.

Since a cooperative movement aims to support and encourage existing employees to educate themselves and improve their skill set by taking multiple pieces of training, it must be considered as a compulsory task for every organization. In order to facilitate the cooperative movement that is set up, this commerce project idea recommends aiding organizations by giving them financial support from cooperative banks.

An Analysis of the Downside of Co-Operative Thrift and Credit Society

Objective:

To show how pointless it is to establish a co-operative and credit society.

We learned from the earlier commerce project idea that Cooperative societies are established to help the employees and save a certain percentage of their salaries. The savings did will come in handy after retirement and can be used for making multiple investments in different areas. However, it is noticed that this type of society doesn’t form a consensus when the goals are not attained. It is seen that these cooperative societies have a disparity in the financial and personal contexts, which is why they are not able to make the employees happy.

These types of things happen because of poor management and the inefficiency of people who are in authority. The irresponsible attitude of people who are in authority killed the interests of its members. Another important aspect of this commerce project idea is that when individuals take loans from cooperative societies, most of the time, they fail to return them thus causing loss of money for the society. When the bad debts increase, the cooperative society gets affected.

Also, Check other Project Ideas

Analyzing the Role of Insurance Companies for Driving Growth of SMEs

Objective:

To thoroughly understand the relationship that exists between SMEs and insurance companies. This project also makes you understand how insurance companies play a big role in the growth of SMEs. This project will involve different studying factors that are responsible for growth as well as failure. One of the major aspects that this project covers is that it talks about the reasons behind SMEs not wanting to insure themselves.

This project will give you an overlook of SMEs and how insurance companies are key to prevent them from getting shut down. SMEs are considered to be small and medium-scale enterprises. A small-scale enterprise consists of almost 5 to 10 employees, whereas a medium-scale enterprise is made of almost 50 to 100 employees.

SMEs are considered important in helping the economy of a country to grow because these are independent enterprises and have the ability to generate effective revenue, which will be treated as a contribution to the country’s GDP. It is good to make sure that these SMEs get covered by insurance. However, due to lack of awareness, many SMEs failed to get covered by insurance which is why they are unable to function for more than two years.

The factors that might cause SMEs to shut down are financial difficulties and departmental challenges in the long run. Insurance companies can help SMEs to overcome these challenges. With an insurance cover, the employees feel secure, which is why they work hard and become more efficient, thus making the SMEs more effective and productive. From this commerce project idea, you will learn that the secret trick to making an SME successfully is insurance cover.

Implications of Globalisation on National Security

Objective:

To examine and evaluate the rate of transnational threats and the methods that are used to breach security. Furthermore, it gives an overall view of the effects of globalization on increased national security.

When you start working on this commerce project idea and deeply analyze it, you will get to know two aspects of globalization. Firstly, it is noticed that wherever there is a threat of security breaching, there is growth in the business. The ease of communication and to and fro of trade are considered threats to national security.

At the same time, it has also helped many organizations to grow their business to a larger extent. The threats of security breaching force organizations to strengthen their security measures and streamline all of their services. Considering the advantages and disadvantages, we conclude that globalization is beneficial for most sectors.

Globalization also helped many underdeveloped countries to improve their national income and stabilize their financial conditions. On top of that, with the help of globalization, industries were able to do a lot of awareness among customers that stimulated the customers to buy items that are needful to them. Globalization is believed to have a positive impact on the national income of a country.

Exploring the Importance of Commerce in Today’s World

Objectives:

To understand the benefits and advantages of commerce in today’s world.

When you start working on this commerce project idea, you will come to know about different aspects of commerce and its benefits. Some of which are:

  • Commerce will help humans to accomplish many of their wants. Trade between different countries was possible because of commerce. When trade or movement of goods takes place, it automatically fulfill the want that humans have. This also puts a positive impact on social welfare. This distribution of goods through e-commerce has helped many small businesses to survive and grow in the competitive market.
  • Commerce helps in improving the standard of living. When the flow of money increases, it automatically makes people buy more things and improve their standard of living as their financial stability becomes better. Product delivery to diverse locations also made it possible for many people to buy things from the comfort of their homes.
  • Commerce empowers both consumers and producers. Imagine, if manufacturers keep producing products and no one is there to buy them, what will happen to the economy? The country’s economy will go into recession in such situations, but commerce prevents such events from happening. Retailers and wholesalers are the ones who facilitate the movement of goods in the market. Now, with the introduction of e-commerce, it has become easier to link customers with retailers and wholesalers.
  • Commerce is responsible for growth in manufacturing, transporting, warehousing, banking, and advertising. So to ensure that these departments run smoothly, organizations require human resources, thus increasing the employment opportunities for people.
  • When trade increases in a country, it has a positive impact on production and consumption rate. When the production and consumption rate increases, it automatically increases the national income of the country.
  • Commerce is responsible for growth in different auxiliary sectors such as insurance, publicity, banking, advertising, and marketing.
  • Commerce has given rise to globalization. International trade has now become a normal thing. Countries are strongly involved in exporting and importing different goods. This also has a positive impact on the relationships of different countries.
  • Commerce has facilitated underdeveloped countries to export their goods to other countries. Such trade increases their flow of money and helps them to develop.

Importance of E-commerce in Emerging Markets

Objective:

To know the benefits of e-commerce, upcoming trends and identify solutions for the present challenges.

You are aware of the fact that the importance of the internet has increased in this couple of years. People across the globe are using the internet and spending most of their time on it. With the rise of the internet, online shopping also got introduced, and it enabled people to buy anything from anywhere. Considering the present situation, it is believed that e-commerce is the future. E-commerce is taking over traditional commerce because it costs you less and is easy to handle.

E-commerce platforms allow consumers to conduct a comparative analysis and buy the product which is available at the best price. When a consumer succeeds to buy a product at the best available price, it is like an achievement for them as they were able to save a penny from their purchase. The clarity in communication between the consumer and the e-commerce platforms makes things easier.

Apart from all the benefits, there are some risks involved with e-commerce. One of the main threats that e-commerce involves is the breaching of customers’ trust. The failure of e-commerce business is possible when everything doesn’t go according to the plans formulated. The only key to achieve success in e-commerce is customer satisfaction.

The upcoming trends in e-commerce will make it possible for business ventures to experience growth in the market. Businesses will start adapting AI technology and augmented reality for improving the customer’s experience. Technology will help e-commerce companies to increase their reach and scalability.

Wrapping Up

As of now, every country’s economy and development depends upon commerce which is why this is the most talked-about field. If you want to be a skilled professional in this field then you must work on any of the above commerce project ideas. After completing your work on these topics, you will have a better knowledge of commerce and how it works.

Interesting Artificial Intelligence Project Ideas and Topics for Beginners

Artificial Intelligence Project Ideas & Topics for Beginners

Artificial Intelligence is one of the most in-demand skills in the job market, with experts being in high demand. Aspiring professionals should start honing their AI skills if they want to be prepared for when it hits the mainstream. Fortunately, there are several ways to learn about AI and experiment with it, even without a degree in computer science. Not only will you have an easier time getting a job later on you may even be able to create your own!

Practical Projects on Artificial Intelligence for Students will help you gain valuable knowledge in this area while knowing the uses AI has in many different sectors will help you see how applicable it is to real life. The more research and experimentation you do on your own, the better grasp you’ll have of the wide variety of different uses of AI.

In this article, we’re going to explore some of the Best Artificial Intelligence Project Ideas & Topics for Beginners that you can use as an ultimate guide to AI. These projects will help you gain hands-on experience and understanding of AI, so by the time you’re done, you’ll be able to put your knowledge to the test. Try out a few of these AI projects and dive deeper into this field.

What is Artificial Intelligence?

Artificial intelligence, or AI, is the capacity of machines to perform human-like tasks commonly associated with thinking or learning. This includes learning, reasoning, and self-correction, which are critical components of human learning, memory, and decision-making.

It can solve problems and learn from experience, allowing it to adapt instead of just following specific rules. Artificial intelligence is slowly becoming part of our daily lives, as you can see with virtual assistants like Siri and Alexa or Chatbots that help customers resolve issues within an e-commerce site.

As AI becomes increasingly pervasive, businesses should be wary of vendors who use the label lightly. Although artificial intelligence sounds like a catch-all term, it’s not — and businesses should avoid using it at all costs. Rather than use AI as an umbrella term for everything machine learning, companies should consider the specifics of their business and product and what machine learning can do for them.

Advantages of AI Projects

Artificial intelligence gives a huge range of benefits to companies, from making businesses more efficient to improving service for customers. AI also helps companies with planning, organizing, and analyzing data. This allows businesses to make better decisions about things like marketing strategy or buying new equipment.

      • Artificial Intelligence Project Ideas allow you to check out new things and learn how they work. If you want to know more about the real-life applications of AI, then it’s one way to do that.
      • With the increasing demand for AI-powered applications, the number of opportunities in this field is bound to increase.
      • By completing an AI project yourself, you’ll understand the technology better and may market your skills to potential employers.
      • Artificial intelligence relies on not only understanding how to use algorithms but also requires you to create original projects with AI tools that haven’t been attempted before.

Projects in AI are a good way to show off your skills. When discussing the specifics of your AI project, be sure to emphasize any creative solutions you proposed or unique challenges that you faced. Here is a list of Artificial Intelligence Project Ideas & Topics for Beginners to work on:

You can also see our Artificial Intelligence Handwritten Notes to prepare the concepts of AI with ease.

List of AI Projects for Beginners to Work on in 2021

1. Stock Price Prediction

Stock Price Prediction is the first on the list of the best Artificial Intelligence Project ideas for beginners. The stock market is loved by machine learning scientists because of its rich data set. One can have access to a wide array of datasets and work with these datasets will help you learn how to use machine-learning algorithms to analyze massive amounts of information.

Students with an interest in finance would be excited by the prospect of using artificial intelligence to predict the fluctuations in the stock market. A large amount of data that’s constantly available makes it easy for these students to find patterns and models that can help them make accurate predictions on the future state of the stock market.

2. Chatbots

Another important and intriguing project in AI is creating a Chatbot. The chatbot could be used as a customer service representative. It’s important to start simple, so the first task would be to make a basic version of the chatbot that could answer questions from users about your business.

With the help of Artificial Intelligence, you can not only create a chatbot for one purpose but create several chatbots for multiple purposes. This will allow you to appeal to different kinds of customers.

3. Enron Investigation

Enron was once the 7th largest company in the United States, but it collapsed in 2000. The collapse was caused by the company’s fraudulent accounting practices, which were uncovered by an investigation conducted by Enron’s board of directors.

Enron has turned into a cautionary corporate tale, underscoring the importance of good business ethics. Unfortunately, the company’s database lives on with next to no security. The database includes records of hundreds of thousands of emails between employees and executives that are now public.

Some messages are dated as recently as 2006, showing that Enron was still operating after it filed for bankruptcy. If you’re working on a project that requires mapping social networks, you can use this database for social network analysis or anomaly detection.

This is probably one of the most popular AI projects to date. The data scientist community is so strong here that users can often find answers to their questions in the forums.

4. Customer Advice

E-commerce giants like Amazon have invested heavily in artificial intelligence. After collecting a user’s data and their behaviors, these systems suggest an item that the user is likely to buy. Just as you might bypass a customer service rep when you see a crowd in a store, these systems help e-commerce sites make for a better shopping experience.

You can use a customer advice system to get immediate feedback on what customers like and don’t like about your product. This works well for E-commerce sites or any business that wants to get feedback from their clients.

You could start this process by attempting to build a real-time messaging tool into your e-commerce application — this would allow you to interact with customers and learn more about what they like and dislike about the product. This is a great way to start your project in the field of Artificial Intelligence.

See More:

Advanced Stage Artificial Intelligence Project Ideas

5. Facial Emotion Recognition and Detection

Facial expression recognition is a trending AI research topic whose goal is to accurately identify and classify facial expressions. This technology can be useful in a variety of situations, such as understanding customer responses to marketing campaigns and working with other human service professionals to make the best decisions about how to respond to their clients.

Facial expression recognition technology has been around for decades, but only recently have researchers shown tangible results of Deep Learning technology applied to this area. Facial recognition software has become accessible to the masses, but this hasn’t stopped software developers from constantly coming up with innovative ways to make facial identification easier.

You can easily track emotions and detect people’s moods, all without the need for them to even be aware of it. Emotion monitoring and recognition can provide a rich experience that focuses on each person’s unique set of emotions and changes daily habits and lifestyles by pinpointing and then retraining our patterns of emotion.

6. Personality Prediction System

The personality Prediction System is one of the fascinating projects in the AI concept. With so many CVs to sift through, it can be difficult for recruiters and managers to find the right candidate — that is, unless we train an AI system to do the dirty work.

By analyzing and interpreting a candidate’s CV, we can improve the quality of our selection process and bring more potential employees into the fold. A personality test built into a candidate management system can help businesses make better hiring decisions.

An intuitive application will allow candidates to input their information and take a personality quiz, which is designed to help determine who is most suited for a job. The tool can be used in the early stages of the hiring process — helping find candidates who are best matched with a company’s mission and values.

The platform’s skill-based matching system ranks users based on their skills and expertise for a given job profile. While traditional platforms consider all different types of essential features, like soft skills or personal patterns, the platform keeps only the most suitable candidates for a given job role. With high precision, it identifies the right person for you within minutes.

7. Online Assignment Plagiarism Checker

Plagiarism is on the rise, but there are smart ways to fight back. By protecting yourself with an AI-powered tool that can detect the originality of your content, you can keep track of your work, spot duplicate content, and boost your SEO. It’s one of the most important ways to stay ahead of the curve in modern writing.

In doing projects in AI in online assignment plagiarism checker, you will develop a plagiarism detector that can detect the similarities in text copies and the percentage of plagiarism. This software uses the text mining method to analyse text and determine its similarity to other texts on an index of several known texts. To create a new account, you must enter your e-mail address and password.

When you upload your essay, the checker scans the document for grammatical errors. Then, it records all of your sources, including citation details and a link to searchable copies of the resource. Once the checker has finished proofreading your paper, it records a grade.

8. Banking Bot

With the prevalence of artificial intelligence in so many fields, banking is one industry that could really use a boost from AI. AI has the power to cut down on manual work and red tape, to check and analyze data quickly and accurately, and to provide users with a more streamlined experience.

If your business is looking for a way to improve its operations, try implementing AI heavily into your current functions. AI for Banks lets you ask questions about your bank account and loan in a natural voice, so you can receive information in real-time.

With AI for Banks, you can easily ask about your bank accounts and loans without lifting a finger — just speak the question aloud and the software will respond on its own. The Banking Bot is an Android application that leverages the power of machine learning to identify users’ needs and projects in AI in banking bot will surely help learners to take their skills to the next level.

Like a chatbot, it communicates with users to get them the information they are looking for but does so using the power of natural language processing. The bot is helpful and communicative — like a human.

9. Heart Disease Prediction Project

This AI-driven application will help patients suffering from heart diseases. Doctors aren’t always available to give feedback and advice, but with this application, they can access guidance and direction whenever they need it by accessing the system remotely.

Patients will be able to track their progress and reach milestones easily. This application will allow patients to get instant access to the consultation and services of certified medical professionals on matters related to heart diseases. Registered users will be able to share and mention their heart-related issues on the online portal and receive customized advice and guidance from doctors based on their needs and requirements.

The system will then process that information to check the database for a specific disease a user might have. This intelligent system uses data mining techniques to guess what a user might have based on the specifics of their symptoms and current conditions. If a condition is found, doctors can be consulted on their opinion on how best to treat the condition, leading to faster diagnoses and better efficiency in healthcare.

Conclusion

In this article, we’ve covered Artificial Intelligence Project Ideas and Topics you can take on to help you learn to use AI in your own work. We started with the basics and finished with some more advanced projects — perfect for any developer looking to build their knowledge of AI.

Mastering artificial intelligence is easier than you think. With the right materials, you can learn everything there is to know about AI in a snap. And once you’re done with these projects, you’ll have the experience necessary to take your coding skills into the next gear.

BTech 3rd Year Metallurgical Engineering Notes PDF | JNTUH & JNTUK BTech III Year Metallurgical Engineering Syllabus, Reference Books

Btech 3rd Year Metallurgical Engineering Notes

BTech Metallurgical Engineering Third Year Notes PDF Free Download: Aspirants who are wondering regarding the Btech Metallurgical Engineering 3rd Year Lecture Notes & study material pdf can refer to this page completely. Here, you can find the better preparation resources along with the III Yr I & II Semester Btech Metallurgical Engineering Handwritten Notes PDF.

Aid your preparation with this best solution ie., Bachelor of Technology Third Year Metallurgical Engineering Lecture Notes by downloading them through accessible links prevailing here. You can either view these Btech Metallurgy 3rd Yr Notes, study materials online/offline with the help of PDF format. So, learn well and score good grades in the exams.

Download BTech Metallurgical Engineering 3rd Year Sem-Wise Notes & Study Materials PDF Free

Get BTech Metallurgical Engineering Third Yeat Subject-wise Notes Pdf in this section and ace up your preparation for sem exams. Along with the Lecture notes pdf download, you may also find the Bachelor of Technology Metallurgy Engineering Third Year Syllabus, Referenece Books List, etc. for self-study and better preparation. These study notes of Btech 3rd Year Metallurgical Engineering helps students to download offline easily and prepare well during their learnings.

JNTUH BTech Third Year Metallurgical Engineering Syllabus

Aspirants are recommended to check out the Jntuh Btech Metallurgical Engineering 3rd year Syllabus from here and grasp all subjects covered in Sem-1 & Sem-2. Also, you can view JNTUK & other institutes’ Bachelor of Technology in Metallurgy Engineering Third Year Syllabus list via the below sections.

III Year I Semester:

  • Management Fundamentals for Engineers
  • Professional Elective – I
  • Fuels, Furnaces & Refractories
  • Nanomaterials
  • Computational Materials Engineering
  • Mechanical Metallurgy
  • Iron Making and Steel Making Technologies
  • Materials Processing – I
  • Mechanical Metallurgy lab
  • Materials Processing Lab -I
  • Fuels Lab

III Year II Semester:

  • Open Elective – I
  • Engineering Materials
  • Metallurgy for Non-Metallurgists
  • Non-Ferrous Extractive Metallurgy
  • Materials Processing – II
  • Environmental Degradation of Materials
  • Professional Elective – II
  • Advanced Iron and Steel Making
  • Composite Materials
  • Electronic Materials
  • Materials Processing Lab -II
  • Advanced English Communication Skills Lab
  • Environmental Degradation of Materials Lab
  • Indian Constitution

JNTUK Bachelor of Technology III Year Metallurgical Engineering Syllabus

III Yr I Sem:

  • Foundry Technology
  • Steel Making
  • Managerial Economics and Financial Analysis
  • Phase Transformations and Heat Treatment
  • Material Testing and Evaluation
  • Foundry Technology Lab
  • Heat Treatment Lab
  • Material Testing and Evaluation Lab
  • IPR & Patents

III Yr II Sem:

  • Materials Joining Techniques
  • Industrial Engineering and Management
  • Materials Characterization
  • Metal Forming
  • OPEN ELECTIVE
  1. Waste Water Management
  2. Robotics
  3. DBMS
  4. CAD/CAM
  5. Functional Materials
  6. Metallurgical Analysis
  • Materials Joining Lab
  • Metal Forming Lab
  • Materials Characterization Lab
  • Professional Ethics & Human Values

Some of the Major Universities Prescribed BTech Metallurgica Engineering 3rd Year Syllabus

SEMESTER 5 SEMESTER 6
Deformation and Fracture Behaviour of Materials Steel Making
Physics Metallurgy- I Foundry Metallurgy
Iron Making Extraction of Non-ferrous Metals
Chemical Kinetics and Mass Transfer Solid-State Phase Transformation Processes
Electro-Chemistry and Corrosion Physics of Metals

List of Suggested Books for Btech Metallurgical Engineering III Year Course Subjects

Here is the list of recommended books for Btech 3rd year metallurgical engineering course subjects, that students may get help from during their preparation.

  • Metals Handbook Vol. 5 published by ASM, Ohio’
  • Making Shaping and Treating of Steels by United States Steel Corporation, Pittsburgh.
  • Dr. B. Kuberudu and Dr. T. V. Ramana: Managerial Economics & Financial Analysis, Himalaya Publishing House, 2014.
  • Heat Treatment of metals-Zakharv-Mir Publishers
  • Engineering Materials Science – CW Richards
  • Intellectual Property Rights (Patents & Cyber Law), Dr. A. Srinivas. Oxford University Press, New Delhi.
  • JF Lancaster: Welding Metallurgy
  • Operations Management by J.G Monks, McGrawHill Publishers
  • Modern Metallographic Techniques & their application – victor Phillips
  • Kurt Lange “Handbook of Metal Forming”, Society of Manufacturing Engineers. Michigan, USA, 1988
  • Environmental Engineering-II: Sewage disposal and Air pollution Engineering, by Garg, S.K., Khanna publishers
  • Robotic Engineering / Richard D. Klafter, Prentice Hall
  • Database Systems design, Implementation, and Management, Peter Rob & Carlos Coronel 7th Edition.
  • Mel Schwartz, “ Encyclopedia of Smart Materials”, Vol. I, John Wiley and Sons, USA, 2002
  • Agarwal, B.C. and Jain S.P., A Text Book of Metallurgical Analysis, Khanna Publishers, Delhi -1963

Metallurgical Engineering III Year Reference Books

Metallurgy and Materials Important Questions List

  • What is a Metallic Bond? How does the type of bonding influence the properties of crystals?
  • Explain Gibbs Phase Rule and its Importance?
  • Describe Ionic Bond, Covalent Bond, Metallic Bond?
  • Draw the unit cells of BCC, FCC, HCP crystal structures?
  • Explain the importance of equilibrium diagrams in the development of new alloys?
  • Define peritectic, eutectoid, and eutectic reactions?
  • Explain electron compounds?
  • What are intermetallic compounds?
  • Name the allotropic forms of iron and explain the lattice structure of each?
  • Draw the iron-iron carbide equilibrium diagram and label all the regions?
  • What are carbides?
  • Explain various nitrides?
  • Discuss applications of c-c composites?
  • Write briefly about thermoplastics and thermosetting plastics?
  • Explain the manufacture of Fibre reinforced plastic and Bring out the typical applications of FRP?

Metallurgy and Materials Important Questions

FAQs on BTech Metallurgical Engineering Third Year Notes Free PDF

1. How do I study for Btech Metallurgy and Materials Notes for 3rd Yr Semester Exams?

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Final Words

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Permutation and Combination Notes | Syllabus, Reference Books and Important Questions

Permutation and Combination Notes

Permutation and Combination Notes: Candidates desiring to have a good grasp of mathematics in a comprehensive manner can obtain notes of Permutation and Combination to understand the topic precisely. The Permutation And Combination Notes comprise a comprehensive study plan and all-important information. Students will get information about the latest Reference Books, Syllabus, and Important Questions List for Permutation And Combination Notes.

The Permutation and Combination Notes Pdf are one of the essential study materials that can improve the students’ preparation for the exam. Students can score better marks with the help of these notes. Students can refer to the Big Data Lecture Notes For CSE as per the latest and updated syllabus from this article.

The article given below helps the students access the best Permutation And Combination Notes Pdf as per the latest curriculum.

Introduction to Permutation and Combination

The process of arranging all the elements of a set into some sequence or order is known as permutation. Permutations transpire, in more or less prominent forms, in almost every area of mathematics. They often arise when several orderings on certain finite sets are considered. The process of selecting items from a collection, such that the order of selection does not matter is known as combination. It is likely possible to count the number of combinations. Both the Permutation and Combination concepts are a fundamental part of Mathematics.

Permutation and Combination Notes PDF and Study Material – Free Download

Permutations and Combinations is an important chapter in Class 11 Maths and you can refer to our Notes to score well in exams. In Permutation And Combination Notes Pdf, you will learn about the basic concepts of the subject and different aspects of the topic. The study materials and notes help the learner to become more active so that the learning process is developed. The students can refer to and use the Permutation And Combination Notes Pdf and Study Materials as a reference. Students pursuing graduation can also download PDF notes. Students can refer to the following formats for Notes for Permutation And Combination and increase their efficiency and problem-solving skills to achieve excellent grades in the exams.

  • Permutation and Combination PDF
  • Permutation and Combination Notes PDF
  • Permutation and Combination Lecture notes PDF
  • Permutation and Combination Handwritten Notes PDF
  • Permutation and Combination Question bank with answers PDF
  • Permutation and Combination of previous year PDF papers
  • Permutation and Combination Notes PPT

Permutation and Combination Definition

Permutation: A permutation is defined as the arrangement of a set of members or objects into a sequence or order. Order matters in case of Permutations.

Combination: Combination is a way of choosing items from a collection and order doesn’t matter here.

Permutations and Combinations Formulas

Permutation Formulas: The number of an arrangement of n objects taken r at a time, where 0 < r ≤ n, denoted by nPr is given by (n!) / (n-r)!. 

  • The number of Permutations of n different objects taken r at a time is given by p1 + p2 + p3 + … + pk = n = n!/p1!p2!p3!..pk!
  • Number of permutation of n different objects taken r at a time, When a particular object is to be included in each arrangement is r. n-1Pr-1
  • When a particular object is always excluded, then a number of arrangements = n-1Pr.
  • The number of permutations of n different objects taken all at a time when m specified objects always come together is m! (n – m + 1)!.
  • The number of permutations of n different objects taken all at a time when m specified objects never come together is n! – m! (n – m + 1)!

Combination Formulas: r things can be made to choose from n things all at a time in a combination then the formula is nCr= (nPr )/( r!)

Relationship between  Permutation and Combination is nCr = nPr/r

Permutation and Combination Reference Books

Reference books for Permutation And Combination Notes are an essential source of information. The reference books provide necessary information and explanation about the topics. The books referred by subject experts help students to understand the subject accurately. Candidates will understand the topics if they understand the updated syllabus thoroughly. Here is a list of the best-recommended books for Permutation And Combination Notes.

  • Essential Permutations and Combinations by Tim Hill
  • Permutation and Combinations by Chandra Ramesh
  • Principles and Techniques in Combinatorics by KOH KHEE MENG
  • Mastering Permutations and Combination: Probability by Duo Code
  • The Ultimate Beginner’s Guide to Permutations & Combinations: Probability by Arthur Taff
  • Choice and Chance, with one thousand exercises by William Allen Whitworth
  • Combination, Permutations, Probabilities by A. Nicolaides
  • Schaum’s Outline of Combinatorics by V. Balakrishnan
  • Introductory Combinatorics by Kenneth P. Bogart

Permutations and Combinations Reference Books

Permutation and Combination Notes PDF Syllabus

The best way to commence your preparation for the Mathematics course is to understand the Permutation And Combination syllabus and the topics of the subject. The syllabus of Permutation And Combination is an important part of the Mathematics course.  The article on Permutation And Combination Notes provides a detailed structure along with the latest and updated syllabus, keeping in mind every student’s requirements and assessing their preparation efficiency. It is a reliable course planning tool that plans and organizes the subject for a student.

The Permutation And Combination Notes is an integral part of the maths curriculum that provides the students a clear idea about what and how to study the subject. The article on Permutation And Combination Notes presents all the essential topics under this section so that students can allot time for each topic and prepare accordingly. Students must cover all the topics before attempting the Maths exam so that the exam paper is easy to answer. The updated syllabus also assures that students remain aware of the Permutation And Combination Notes updated syllabus to prevent wasting unnecessary time on irrelevant topics.

Here is a list of the updated syllabus for Permutation And Combination Notes

Topics Description
Fundamental Principle of Counting
  • Multiple Principle of Accounting
  • Additional Principle of Accounting
Permutation
  • Counting Formula for Permutation
  • N Distinct object Permutation when repetition is allowed
  • N arrangement of things when all are not distinct
Circular Permutation
  • Different Things Taken r at a time and  number of circular permutations of n
Combinations
  • Counting Formula for Combinations
  • Identical and Distinct Objects Selections
  • Arrangements in Groups

List of Permutation and Combination Important Questions

Candidates pursuing mathematics can refer to the list of all the essential questions of Permutation and Combination stated below. Here is a list of questions that are aimed to help the students to excel in the examination and understand the subject precisely.

  • How many different words can be created, using all the letters of the word GIFT?
  • Find out the three-digit numbers that can be created applying all the digits of 2, 3, and 4.
  • What are the other words the letters of the word MAGIC are made?
  • What are permutations? Give examples.
  • What are some of the real-life examples of combinations and Permutations?
  • Write the relation between permutations and combinations.
  • Give examples of the difference between permutations and combinations?
  • What is the formula of Permutation and Combinations?

Permutation and Combination Important Questions

Frequently Asked Questions on Permutation and Combination Notes

Question 1.
What is Permutation?

Answer:
The process of arranging all the elements of a set into some sequence or order is known as permutation. Permutations transpire, in more or less prominent forms, in almost every area of mathematics.

Question 2.
Define the term Combination.

Answer:
The process of selecting items from a collection, such that the order of selection does not matter is known as combination.

Question 3.
How does the syllabus of Permutation and Combination help students for their exam?

Answer:  
The syllabus of Permutation And Combination is a reliable course planning tool that plans and organises the subject for a student. It presents all the essential topics so that students can allot time for each topic and prepare accordingly.

Question 4.
Name some of the reference books for Permutation and Combination.

Answer:
Some of the reference books for Permutation and Combination are:

  • Permutations, Combination, Probabilities
  • Schaum’s Outline of Combinatorics
  • Permutation and Combinations
  • Introductory Combinatorics

Conclusion

The Permutation And Combination Notes PDF and Study Materials presented above are aimed to assist the students at the time of exam preparations. They have authoritative references and are reliable and focus on helping graduates and improving their knowledge and understanding of the subject during the time of preparation of the exam. Students can refer and practice from the provided notes for Permutation And Combination Notes and essential questions from this article.

Probability Questions and Answers | Computer Science Quiz

Computer Science Probability MCQ Quiz Questions and Answers PDF Download

Probability Questions and Answers | Computer Science Quiz

Question 1.
Let A and B be any two arbitrary events, then, which one of the following is true?
(a) P (A ∩ B) = P(A) P(B)
(b) P (A ∪ B) = P(A) + P(B)
(c) P (A | B) = P(A ∩ B)/P(B)
(d) P(A ∪ B)< P(A) + P(B)

Answer/Explanation

Answer:(c) P (A | B) = P(A ∩ B)/P(B)
Explanation:
(a) P(A ∩ B) = P(A) P(B) is false since this is true if and only if A and B are independent events.
(b) P (A ∪ B) = P(A) + P(B) is false since P(A ∩ B) is zero if and only if A and B are mutually exclusive.
(c) P (A | B) = P(A ∩ B)/P(B) is true.
(d) P (A ∪ B) < P(A) + P(B) is false.
Since P(A ∪ B) ≤ P(A) + P(B)


Question 2.
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 4 }{ 5 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 3 }\)

Answer/Explanation

Answer: (c) \(\frac { 1 }{ 2 }\)
Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 1

The bag contains 10 white balls and 15 black balls. Required probability

Probability Questions and Answers Computer Science Quiz chapter 5 img 2


Question 3.
Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is
(a) \(\frac { 1 }{ 36 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 25 }{ 36 }\)
(d) \(\frac { 11 }{ 36 }\)

Answer/Explanation

Answer:(d) \(\frac { 11 }{ 36 }\)
Explanation:
P(atleast one of dice will have 6 facing = 1 – P(none of dice have 6 facing up) = 1 – [\(\frac { 5 }{ 6 }\) × \(\frac { 5 }{ 6 }\)] = 1 – \(\frac { 25 }{ 36 }\) = \(\frac { 11 }{ 36 }\)


Probability Questions and Answers | Computer Science Quiz

Question 4.
The probability that top and bottom cards of a randomly shuffled deck are both aces is
(a) \(\frac { 4 }{ 52 }\) x \(\frac { 4 }{ 52 }\)
(b) \(\frac { 4 }{ 52 }\) x \(\frac { 3 }{ 52 }\)
(c) \(\frac { 4 }{ 52 }\) x \(\frac { 3 }{ 51 }\)
(d) \(\frac { 4 }{ 52 }\) x \(\frac { 4 }{ 51 }\)

Answer/Explanation

Answer:(c) \(\frac { 4 }{ 52 }\) x \(\frac { 3 }{ 51 }\)
Explanation:
The probability that the bottom card of a randomly shuffled deck is ace = \(\frac { 4 }{ 52 }\) because there are 4 aces out of total 52 cards. From the remaining 3 aces out of 51 cards the probability that the top card is also an ace = \(\frac { 3 }{ 51 }\). So required probability = \(\frac { 4 }{ 52 }\) × \(\frac { 3 }{ 51 }\).


Question 5.
The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?
(a) 0.3
(b) 0.25
(c) 0.35
(d) 0.4

Answer/Explanation

Answer:(d) 0.4
Explanation:
P(Rain today) = 0.5
P(Rain tomorrow) = 0.6
P(Rain today ∪ Rain tomorrow) = 0.7
P(Rain today ∩ Rain tomorrow) = ?
P(rain today ∩ Rain tomorrow) = P(Rain today) + P(rain tomorrow) – P(Rain today ∩ Rain tomorrow)
So; 0.7 = 0.5 + 0.6 – P(Rain today ∩ Rain tomorrow) P(Rain today ∩ Rain tomorrow) = 0.5+ 0.6-0.7 = 0.4
So, the probability that it will rain today and tomorrow is 0.4.


Question 6.
A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is
(a) \(\frac { 1 }{ 6 }\)
(b) \(\frac { 3 }{ 8 }\)
(c) \(\frac { 1 }{ 8 }\)
(d) \(\frac { 1 }{ 2 }\)

Answer/Explanation

Answer:(b) \(\frac { 3 }{ 8 }\)
Explanation:
Probability of getting an odd number in rolling of a die = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\). Now using binomial distribution P(Exactly one odd number among three outcomes)
= 3C1( \(\frac { 1 }{ 2 }\))1 ( \(\frac { 1 }{ 2 }\))2 = 3 × ( \(\frac { 1 }{ 2 }\))3 = ( \(\frac { 3 }{ 8 }\))


Probability Questions and Answers | Computer Science Quiz

Question 7.
Suppose that the expectation of a random variable X is 5. Which of the following statements is true?
(a) There is a sample point at which X has the value 5.
(b) There is a sample point at which X has a value greater than 5.
(c) There is a sample point at which X has a value greater than or equal to 5.
(d) None of the above

Answer/Explanation

Answer:(c) There is a sample point at which X has a value greater than or equal to 5.
Explanation:
If all the points have X < 5, then the expectation of a random variable X is surely less than 5. So according to this, there should be at least a sample point at which X ≥ 5.


Question 8.
Consider two events E1 and E2 such that probability of E1 Pr [E1] = \(\frac { 1 }{ 2 }\), probability of E2, Pr [E2] = \(\frac { 1 }{ 3 }\), and probability of E1 and E2, Pr[E1 and E2] = \(\frac { 1 }{ 5 }\) . Which of the following statements is/are true?
(a) Pr[E1 or E2] is \(\frac { 2 }{ 3 }\)
(b) Events E1 and E2 are independent
(c) Events E1 and E2 are not independent
(d) pr \(\left[\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right]\) = \(\frac { 4 }{ 5 }\)

Answer/Explanation

Answer:(c) Events E1 and E2 are not independent
Explanation:
P(E1)= \(\frac { 1 }{ 2 }\) , P(E2) = \(\frac { 1 }{ 3 }\) and P(E1 ∩ E2) = \(\frac { 1 }{ 5 }\)

(a) P(E1 or E2)
= P(E1) + P(E2) – P(E1 ∩ E2)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\) – \(\frac { 1 }{ 5 }\) = \(\frac { 19 }{ 30 }\)
However given in option (a) is \(\frac { 2 }{ 3 }\).
So option (a) is not true.

(b) For independent events P(E1 ∩ E2) = P(E1) P(E2)
Here, P(E1 ∩ E2) = \(\frac { 2 }{ 5 }\)
P(E1) P(E2) = \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 3 }\) = \(\frac { 1 }{ 6 }\)
So; (P E1 ∩ E2) ≠ P(E1) P(E2)
So events E1 and E2 are not independent. Option (b) is not true.

(c) Since E1 and E2 are not independent So option (c) is true.

(d) P(E1/E2) =\(\frac{\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{2}\right)}\) = \(\frac { 3 }{ 5 }\) So option (d) P(E1/E2) = \(\frac { 4 }{ 5 }\) is false.


Question 9.
E1 and E2 are events in a probability space satisfying the following constraints:

  • Pr(E1) = Pr(E2)
  • Pr(E1 ∪ E2) = 1
  • E1 and E2 are independent

The value of Pr(E1), the probability of the event E1, is
(a) 0
(b) \(\frac { 1 }{ 4 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) 1

Answer/Explanation

Answer:(d) 1
Explanation:
Constraints are
(i) P(E1) = P(E2) = x
(ii) P(E1 ∪ E2) = 1
(iii) E1 and E2 are independent so
P(E1 ∩ E2) = P(E1) P(E2) = x × x = x2

NOW,
P(E1 ∪ E2) = P(E1)+P(E2) – P(E1 ∩ E2)
1 = x + x – x2
1 = 2x – x2
x2 – 2x + 1 = 0
(x – 1)2 = 0
x = 1
So; P(E1) = P(E2) = x = 1


Probability Questions and Answers | Computer Science Quiz

Question 10.
Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
(a) \(\frac{1}{7^{7}}\)
(b) \(\frac{1}{7^{6}}\)
(c) \(\frac{1}{2^{7}}\)
(d) \(\frac{7}{2^{7}}\)

Answer/Explanation

Answer:(b) \(\frac{1}{7^{6}}\)
Explanation:
Sample space = 77
All accidents on the same day = 7 ways (all on Monday, all on Tuesday…)
So, required probability = \(\frac{7}{7^{7}}\) = \(\frac{7}{7^{6}}\)


Question 11.
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is
(a) \(\frac { 1 }{ 16 }\)
(b) \(\frac { 1 }{ 8 }\)
(c) \(\frac { 7 }{ 8 }\)
(d) \(\frac { 15 }{ 16 }\)

Answer/Explanation

Answer:(c) \(\frac { 7 }{ 8 }\)
Explanation:
p(atleast one head and one tail)
= 1 – p(no head or no tail)
= 1 – [p(no head) + p(no tail) – p(no head and no tail)]
= 1 – [p (all tails) + p(all heads) – 0]
= 1 – [\(\left[\frac{1}{2^{4}}+\frac{1}{2^{4}}-0\right]\) = 1 – \(\frac { 2 }{ 16 }\) = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\).

Or

Alternate Method:
We need atleast one head (≥1H) and atleast one tail (≥ 1 T). First we satisfy ≥ 1H as follows.
1 H, 3T
2 H, 2 T
3H, 1 T
and 4 H, 0 T
now to satisfy the second condition of ≥ 1 T, we have to remove 4 H, 0 T.
So, the favorable cases are only 1 H, 2 H, and 3 H. The probability of this by binomial distribution formula is
= \(\frac{{ }^{4} C_{1}}{2^{4}}\) + \(\frac{{ }^{4} C_{2}}{2^{4}}\) + \(\frac{{ }^{4} C_{3}}{2^{4}}\)= \(\frac { 7 }{ 8 }\)


Question 12.
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, and A and B are independent, the values of P(A | B) and P(B | A) respectively are
(a) \(\frac { 1 }{ 4 }\) , \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 2 }\) , \(\frac { 1 }{ 4 }\)
(c) \(\frac { 1 }{ 2 }\) , 1
(d) 1 , \(\frac { 1 }{ 2 }\)

Answer/Explanation

Answer:(d) 1 , \(\frac { 1 }{ 2 }\)
Explanation:
Given, P(A) = 1
P(B) = \(\frac { 1 }{ 2 }\)
Since both events are independent
P(A|B) = P(A) = 1
P(B|A) = P(B) = \(\frac { 1 }{ 2 }\)


Question 13.
A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by
(a) f1(t) + f2(t)
(b) \(\int_{t}^{0}\) f1 (X) f2 (X)dX
(c) \(\int_{t}^{0}\) f1 (X) f2 (t – X)dX
(d) max {f1(t), f2(t)}

Answer/Explanation

Answer:(c) \(\int_{t}^{0}\) f1 (X) f2 (t – X)dX
Explanation:
Let the time taken for first and second modules be represented by x and y and total time = t.
∴ t = x + y is a random variable. Now the joint density function
g(t) = \(\int_{0}^{t}\)f(x,y)dx
= \(\int_{0}^{t}\)f (x, t – x)dx
= \(\int_{0}^{t}\)f1(x) f2(t – x)dx
which is also called as convolution of f1 and f2, abbreviated as f1 * f2.Correct answer is therefore, choice (c).


Probability Questions and Answers | Computer Science Quiz

Question 14.
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?
(a) \(\frac { 3 }{ 8 }\)
(b) \(\frac { 1 }{ 2 }\)
(c) \(\frac { 5 }{ 8 }\)
(d) \(\frac { 3 }{ 4 }\)

Answer/Explanation

Answer:(a) \(\frac { 3 }{ 8 }\)
Explanation:
Given, P(H) = \(\frac { 1 }{ 2 }\)
P(T) = \(\frac { 1 }{ 2 }\)
Apply Bernoulli’s formula for binomial distribution,
P(X = 2) = \({ }^{4} \mathrm{C}_{2}\)(1/2)2(1 – \(\frac { 1 }{ 2 }\))4 – 2
= \({ }^{4} \mathrm{C}_{2}\) ( \(\frac { 1 }{ 2 }\) )2 (1/2)2
= \(\frac{{ }^{4} \mathrm{C}_{2}}{2^{4}}\) = \(\frac { 6 }{ 16 }\) = \(\frac { 3 }{ 8 }\)


Question 15.
In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?
(a) \(\frac { 3 }{ 23 }\)
(b) \(\frac { 6 }{ 23 }\)
(c) \(\frac { 3 }{ 10 }\)
(d) \(\frac { 3 }{ 5 }\)

Answer/Explanation

Answer:(b) \(\frac { 6 }{ 23 }\)
Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 3

Total number of children = 50 × 3 + 30 × 2 + 20 × 1 = 230
Favorable cases = The number of children who come from 2 children families = 30 × 2 = 60
So the probability that a randomly picked child belongs to a 2 children families = \(\frac { 60 }{ 230 }\) = \(\frac { 6 }{ 23 }\)


Question 16.
Let X and Y be two exponentially distributed and independent random variables with mean a and β, respectively. If Z = min (X,Y), then the mean of Z is given by
(a) (1/(α + β))
(b) min (α, β)
(c) (αβ/(α + β))
(d) α + β

Answer/Explanation

Answer:(c) (αβ/(α + β))
Explanation:
f(x) = λ.e-λx, x ≥ 0
X and Y are two independent exponentially distributed random variables. Let λ1 and λ2 parameters of X and Y respectively.
P(X ≥ x)= e-λx, x > 0
P(Y ≥ x)=e2x, x > 0
Given Z = min (X, Y)
P(Z ≥ x) = P(X ≥ x, Y ≥ x)
= P(X ≥ x) P(Y ≥ x)
= e1x . e2x = e-(λ1 + λ2)x
Since mean of exponential distribution = 1/Parameter
So, α = 1/λ1 ⇒ λ1 = 1/α
β = 1/λ2 ⇒ λ2 = 1/β
∴ Z is random variable with parameter
Mean of Z = IMG


Question 17.
An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches — 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is
(a) 0
(b) 2550
(c) 7525
(d) 9375

Answer/Explanation

Answer:(d) 9375
Explanation:
Let the marks obtained per question be a random variable X. Its probability distribution table is given below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 4

Expected marks per question = E(x)=ΣXP(X)
= 1 × \(\frac { 1 }{ 4 }\) + (-o.25) × \(\frac { 3 }{ 4 }\)
= \(\frac { 1 }{ 4 }\) – \(\frac { 3 }{ 16 }\) = \(\frac { 1 }{ 16 }\) marks
Total marks expected for 150 questions = \(\frac { 1 }{ 16 }\) × 150 = \(\frac { 75 }{ 8 }\) marks per student
Total expected marks of 1000 students = \(\frac { 75 }{ 8 }\) ×1000 =9375 marks
So correct answer is (d).


Question 18.
Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is
(a) nCd/2n
(b) nCd/2d
(c) d/2n
(d) 1/2d

Answer/Explanation

Answer:(a) nCd/2n
Explanation:
If the hamming distance between two n bit strings is d, we are asking that d out of n trials be successful (success here means that the bits are different). So this is a binomial distribution with n trials and d successes and probability of success

P = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(Since out of the 4 possibilities {(0,0), (0,1), (1, 0), (1,1)} only two of them (0,1) and (1,0) are success)
So p(X = d) = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{d}}(1 / 2)^{\mathrm{d}}(1 / 2)^{\mathrm{n}-\mathrm{d}}\) = \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{d}}}{2^{\mathrm{n}}}\)
Correct choice is therefore (a).


Probability Questions and Answers | Computer Science Quiz

Question 19.
A point is randomly selected with uniform probability in the X-Y. Plane within the rectangle with corners at (0, 0), (1, 0), (1, 2) and (0, 2). If p is the length of the position vector of the point, the expected value of p2 is
(a) \(\frac { 2 }{ 3 }\)
(b) 1
(c) \(\frac { 4 }{ 3 }\)
(d) \(\frac { 5 }{ 3 }\)

Answer/Explanation

Answer:
(d) \(\frac { 5 }{ 3 }\)
Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 5

Length of position vector of point = p = \(\sqrt{x^{2}+y^{2}}\)
p2 = x2 + y2
E(p2) = E(x2 + y2) = E(x2) + E(y2)
Now x and y are uniformly distributes 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2 Probability density function of x = \(\frac { 1 }{ 1 – 0 }\) = 1
The probability density function of

Probability Questions and Answers Computer Science Quiz chapter 5 img 6


Question 20.
Let f(x) be the continuous probability density function of a random variable X. The probability that a < X ≤ b, is
(a) f (b – a)
(b) f(b) – f(a)
(c) \(\int_{b}^{a}\) f(X)dX
(d) \(\int_{b}^{a}\) xf (X)dX

Answer/Explanation

Answer:(c)\(\int_{b}^{a}\) f(X)dX
Explanation:
If f(x) is the continuous probability density function of a random variable X then, p(a < x ≤ b) = p(a ≤ x ≤ b) = \(\int_{b}^{a}\)f (x)dx


Question 21.
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
(a) \(\frac { 1 }{ 36 }\)
(b) \(\frac { 1 }{ 6 }\)
(c) \(\frac { 1 }{ 4 }\)
(d) \(\frac { 1 }{ 3 }\)

Answer/Explanation

Answer:
(b) \(\frac { 1 }{ 6 }\)
Explanation:
The given condition corresponds to sampling with replacement and with order. No 2 marbles have the same color i.e. Drawn 3 different marble. So total number of ways for picking 3 different marbles = 3! = 6. Probability of getting blue, green, red in order
= \(\frac { 10 }{ 60 }\) × \(\frac { 20 }{ 60 }\) × \(\frac { 30 }{ 60 }\) × 6
[Since 6 ways to get the marbles] = \(\frac { 1 }{ 6 }\)


Question 22.
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is
(a) 3
(b) 4
(c) 5
(d) 6

Answer/Explanation

Answer:(a) 3
Explanation:
E(X) = S(Xi × Pi)
Where Xi = number of tosses until we get successive HEAD or TAIL.
Pi = Probability that we get in Xi tosses. Total possibilities by tossing a coin two times i.e. HT, HH, TH, TT. Out of which favorable cases are HH, TT.
P(X = 2) = \(\frac { 2 }{ 4 }\)
Similarly for X = 3 only THH and HTT are favorable out of a total of 8 outcomes.
So, P(X = 3) = \(\frac { 2 }{ 8 }\)
Similarly, we can see that in every case we will have only 2 favorable cases and 2n sample space.
So for nth toss probability = P(X = n) = 2/2n So the probability distribution table for X (the number of Tosses) is given below:

X 2 3 4 6
P(X) 2/4 2/8 2/16 2/32

E(X) = 2 × \(\frac { 2 }{ 4 }\) + 3 × \(\frac { 2 }{ 8 }\) + 4 × \(\frac { 2 }{ 16 }\) + …… It is combination of AP and GP form. So multiplying E(X) by \(\frac { 1 }{ 2 }\) and subtracting from E(X)
E(X) = 2 × \(\frac { 1 }{ 2 }\) + 3 × \(\frac { 1 }{ 4 }\) + 4 × \(\frac { 1 }{ 8 }\) + …..
\(\frac { 1 }{ 2 }\) × E(X) = 2 × \(\frac { 1 }{ 4 }\) + 3 × \(\frac { 1 }{ 8 }\) + 4 × \(\frac { 1 }{ 16 }\) + …..
After solving both equations we get,
\(\frac { 1 }{ 2 }\) × E(X) = 1 + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 16 }\) + ……
\(\frac { 1 }{ 2 }\) × E(X) = 1 + \(\frac{1 / 4}{(1-1 / 2)}\) ⇒ E(X) = \(\frac { 6 }{ 4 }\) × 2 = 3


Probability Questions and Answers | Computer Science Quiz

Question 23.
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C.
What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?
(a) 0.4
(b) 0.6
(c) 0.8
(d) 0.9

Answer/Explanation

Answer:(c) 0.8
Explanation:
Let RA : Rain in the afternoon T > 25: Temperature more than 25°C
Let the desired probability = P(RA | T ≤ 25) = x
The tree diagram for this problem is given below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 7

Given P(RA) = 0.6 by rule of total probability P(RA)
= \(\frac { 1 }{ 2 }\) × 0.4 + \(\frac { 1 }{ 2 }\) × x = 0.6 ⇒ x = 0.8


Question 24.
When a coin is tossed, the probability of getting a Head is p, 0 < p < 1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
(a) 1/p
(b) 1/(1 – p)
(c) 1/P2
(d) 1/(1 – p2)

Answer/Explanation

Answer:(a) 1/p
Explanation:
The number of attempts to first succeed follows a geometric distribution. It is well known that the expected value in geometric distribution E(X) = 1/p. Where p is the probability of success in any one attempt.

Alternate Method:

Let X be the number of attempts to first success. Let p be the probability of success in any one attempt. Now the probability distribution table of X is given below:

X 1 2 3 4
P(X) p (1 – p)p (1 – p)2p (1 – p)3p

E(X) = Σ xp(x) = 1 × p + 2 × (p – 1) p + 3 × (p – 1)2 p + ……
This is the arithmetico-geometric series which can be solved as E(X) = 1/p


Question 25.
Suppose there are two coins. The first coin gives heads with probability \(\frac { 5 }{ 8 }\) when tossed, while the second coin gives heads with probability \(\frac { 1 }{ 4 }\). One of the two coins is picked up at random with equal probability and tossed. What is the probability of obtaining heads?
(a) \(\frac { 7 }{ 8 }\)
(b) \(\frac { 1 }{ 2 }\)
(c) \(\frac { 7 }{ 16 }\)
(d) \(\frac { 5 }{ 32 }\)

Answer/Explanation

Answer:(c) \(\frac { 7 }{ 16 }\)
Explanation:
The tree diagram for the problem is given below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 8


Probability Questions and Answers | Computer Science Quiz

Question 26.
Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3,…, 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 10 }\)
(c) \(\frac { 9! }{ 20! }\)
(d) None of these

Answer/Explanation

Answer:(d) None of these
Explanation:
The number of permutations with ‘2’ in the first position = 19!
The number of permutations with ‘2’ in the second position = 10 × 18!
(Fill the first space with any of the 10 add numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
A number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(Fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with the remaining 17 numbers) and so on until ‘2’ is in 11th place. After that, it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations that satisfy the given condition is 19! + 10 × 18! + 10 × 9 × 17! + 10 × 9 × 8 × 16! + … + 10! × 9!

Now the probability of this happening is given by \(\frac{19 !+10 \times 18 !+10 \times 9 \times 17 ! \ldots+10 ! \times 9 !}{20 !}\)
which is clearly not choices (a), (b) or (c)
∴ Answer is (d) none of these.


Question 27.
In a multi-user operating system on average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three, or five requests are made in 45 minutes is given by:
(a) 6.9 × 106 × e-20
(b) 1.02 × 106 × e-20
(c) 6.9 × 103 × e-20
(d) 1.02 × 103 × e-20

Answer/Explanation

Answer:(b) 1.02 × 106 × e-20
Explanation:
The requests follow poisson distribution with α = 20 requests/hr
The observation period = △T = 45 minutes = \(\frac { 45 }{ 60 }\) = \(\frac { 3 }{ 4 }\)hr
So the parameter for the poisson distribution = λ = α △ T= 20 × \(\frac { 3 }{ 4 }\) = 15
The required probability P(X = 1) + P(X = 3) + P(X = 5)
= \(\frac{e^{-15} 15^{1}}{1 !}\) + \(\frac{e^{-15} 15^{3}}{3 !}\) + \(\frac{e^{-15} 15^{5}}{5 !}\) = 1.02 × 106 × e-20


Question 28.
A sample space has two events A and B such that probabilities P(A ∩ B) = 1/2, P( \(\overline{\mathrm{A}}\) ) = 1/3, P( \(\overline{\mathrm{B}}\) ) = 1/3. What is P(A ∪ B)?
(a) \(\frac{11}{12}\)
(b) \(\frac{10}{12}\)
(c) \(\frac{ 9}{12}\)
(d) \(\frac{8}{12}\)

Answer/Explanation

Answer:(b) \(\frac{10}{12}\)
Explanation:
We know that:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\overline{\mathrm{P}(\mathrm{A})}\) = 1 – P(A) and \(\overline{\mathrm{P}(\mathrm{B})}\) = 1 – P(B)
So, P(A) = ( 1 – \(\overline{\mathrm{P}(\mathrm{A})}\)) and P(B) = (1 – \(\overline{\mathrm{P}(\mathrm{B})}\))
P(A ∪ B) = (1 – \(\overline{\mathrm{P}(\mathrm{A})}\)) + (1 – \(\overline{\mathrm{P}(\mathrm{B})}\)) – P(A ∩ B)
= (1 – 1/3) + (1 – 1/3) – (1/2)
= (2/3) + (2/3) – (1/2)
= (4/3) – (1/2) = (5/6) = (10/12)


Probability Questions and Answers | Computer Science Quiz

Question 29.
What is the probability that in a randomly chosen group of r people at least three people have the same birthday?

Probability Questions and Answers Computer Science Quiz chapter 5 img 9

Answer/Explanation

Answer:
(c) 1 – \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\) – rC2 . 365 . \(\frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
Explanation:
P(at least three people have the same birthday) = 1 – P (all have different b’days) – P(exactly two people have same b’day)
Now, P(all have different b’days) = \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\)
P(exactly two people have same b’day) = \({ }^{r} C_{2} \cdot 365 \cdot \frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
[rC2 ways to choose who those two people with same b’day are, 365 ways to choose what the b’day is]
Now,
P (at least three people have the same birthday)
= 1 – \(\frac{365.364 \ldots(365-r+1)}{365^{r}}\)
= \({ }^{r} C_{2} \cdot 365 \cdot \frac{364.363 \ldots(364-(r-2)+1)}{364^{r-2}}\)
Which is option (c).


Question 30.
Aishwarya studies either computer science or mathematics every day. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
(a) 0.24
(b) 0.36
(c) 0.4
(d) 0.6

Answer/Explanation

Answer:(c) 0.4
Explanation:
Let C denote computes science study and M denotes maths study. The tree diagram for the problem can be represented as shown below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 10

Now by the rule of total probability we total up the desired branches (✓) and get the answer as shown below:
p(C on Monday and C on Wednesday)
= p(C on Monday, C on Tuesday and C on Wednesday) + p(C on Monday, M on Tuesday, and C on Wednesday)
= 1 × 0.4 × 0.4 + 1 × 0.6 × 0.4 = 0.16 + 0.24 = 0.40


Question 31.
Let X be a random variable following a normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown. If P(X ≤ -1) = P(Y ≥ 2) the standard deviation of Y is
(a) 3
(b) 2
(c) \(\sqrt{2}\)
(d) 1

Answer/Explanation

Answer:(a) 3
Explanation:
Given µX = 1, σx2 = 4 => σx = 2 and µY = -1.σY is unknow
Given, p(X ≤ -1) = p(Y ≥ 2)
Converting into standard normal variates

Probability Questions and Answers Computer Science Quiz chapter 5 img 11

Now since we know that in a standard normal distribution,
p(z ≤ -1) = p(z ≥ 1) … (ii)
Comparing (i) and (ii) we can say that
\(\frac{3}{\sigma_{\mathrm{y}}}\) = 1 ⇒σy = 3


Probability Questions and Answers | Computer Science Quiz

Question 32.
An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even-numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?
(a) 0.453
(b) 0.468
(c) 0.485
(d) 0.492

Answer/Explanation

Answer:(b) 0.468
Explanation:
It is given that, p(odd) = 0.9p(even)
Now since, Σp(x) = 1
∴ p(odd) + p (even) = 1
⇒ 0.9 p (even) + p (even) = 1
⇒ p(even) = \(\frac { 1 }{ 1.9 }\) = 0.5263
Now, it is given that p (any even face) is same i.e. p(2) = p(4) = p(6)
Now since,
p(even) = p(2) or p(4) or p(6)
= p(2) + p(4) + p(6)
∴ p(2) = p(4) = p(6)
= \(\frac { 1 }{ 3 }\) p (even) = \(\frac { 1 }{ 3 }\) (0.5263) = 0.1754
It is given that
p(even | face > 3) = 0.75
⇒ \(\frac{p(\text { even } \cap \text { face }>3)}{p(\text { face }>3)}\) = 0.75
⇒ \(\frac{\mathrm{p}(\text { face }=4,6)}{\mathrm{p}(\text { face }>3)}\) = 0.75
⇒ p (face>3) = \(\frac{p(\text { face }=4,6)}{0.75}\)
= \(\frac{p(4)+p(6)}{0.75}\)
= \(\frac{0.1754+0.1754}{0.75}\)
= 0.4677 ≃ 0.468


Question 33.
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company, therefore, subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
(a) pq+(1 – p) (1 – q)
(b) (1 – q)p
(e) (1 – p)q
(d) pq

Answer/Explanation

Answer:(a) pq+(1 – p) (1 – q)
Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 12

The tree diagram of probabilities is shown above. From above tree, by rule of total probability, p(declared faulty) = pq + (1 – p) (1 – q)


Question 34.
What is the probability that a divisor of 1099 is a multiple of 1096?
(a) \(\frac{1}{625}\)
(b) \(\frac{4}{625}\)
(c) \(\frac{12}{625}\)
(d) \(\frac{16}{625}\)

Answer/Explanation

Answer:(a) \(\frac{1}{625}\)
Explanation:
p(multiple of 1096 | divisor of 1099) = \(\frac{\mathrm{p}\left(\text { multiple of } 10^{96} \& \text { divisor of } 10^{99}\right)}{\mathrm{p}\left(\text { divisor of } 10^{99}\right)}\) = \(\frac{\mathrm{n}\left(\text { multiple of } 10^{96} \& \text { divisor of } 10^{99}\right)}{\mathrm{n}\left(\text { divisor of } 10^{99}\right)}\)
Since, 10 = 2 . 5
1099 = 299. 599
Any divisor of 1099 is of the form 2a . 5b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99. The number of such divisors is given by (99+1) × (99 + 1) = 100 × 100.
So, no. of divisors of 10″ = 100 × 100.
Any number which is a multiple of 1096 as well as divisor of 1099 is of the form 2a. 5b where 96 ≤ a ≤ 99 and 96 ≤ b ≤ 99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 1096 as well as a divisor of 1099.
∴ p(multiple of 1096 | divisor of 1099)
= \(\frac { 4 × 4 }{ 100 × 100 }\) = \(\frac { 1 }{ 625 }\)


Question 35.
If two fair coins are flipped and at least one of the outcomes is known to be ahead, what is the probability that both outcomes are heads?
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)

Answer/Explanation

Answer:(a) \(\frac{1}{3}\)
Explanation:
Let A be the event of head in one coin. B be the event of head in second coin. The required probability is
P(A ∩ B | A ∪ B) = \(\frac { P{(A ∩ B) ∩ (A ∪ B)} }{ P(A ∪ B }\) = \(\frac { P(A ∩ B }{ P (A ∪ B)}\)
P( A ∩ B) = p(both coin heads) = p(H , H) = \(\frac { 1 }{ 4 }\)
P(A ∪ B) = p(at least one head)
p(HH, HT, TH) = \(\frac { 3 }{ 4 }\)
So required probability = \(\frac { 1/4 }{ 3/4 }\) = \(\frac { 1 }{ 3 }\)


Probability Questions and Answers | Computer Science Quiz

Question 36.
If the difference between the expectation of the square of a random variable (E [x2] and the square of the expectation of the random variable (E [x])2 is denoted by R, then
(a) R = 0
(b) R < 0 (c) R ≥ 0 (d) R > 0

Answer/Explanation

Answer:
(c) R ≥ 0
Explanation:
V(x) = E(x2) – [E(x)]2 = R
where V(x) is the variance of x,
Since variance is σ2x and hence never negative, R ≥ 0.


Question 37.
Consider the finite sequence of random values X = [x1, x2,…, xn]. Let µx be the mean and σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi = a * xn + b, where a and b are positive constants. Let µy be the mean and σy be the standard deviation of this sequence. Which one of the following statements INCORRECT?
(a) Index position of the mode of X in X is the same as the index position of the mode of Y in Y.
(b) Index position of the median of X in X is the same as the index position of the median of Y in Y
(c) µy = aµx + b
(d) σy = aσx + b

Answer/Explanation

Answer:
(d) σy = aσx + b
Explanation:
Standard deviation is affected by scale but not by the shift of origin.
So, yi = axi + b
⇒ σy = aσx
if a could be negative then σy = | a | σx is more correct since standard deviation cannot be negative)
Clearly, σy = aσx + b is false
So (d) is incorrect.


Question 38.
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?
(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 4 }{ 25 }\)
(c) \(\frac { 1 }{ 4 }\)
(d) \(\frac { 2 }{ 5 }\)

Answer/Explanation

Answer:
(a) \(\frac { 1 }{ 5 }\)
Explanation:
The five cards are {1, 2, 3, 4, 5}
Sample space = 5 × 4 ordered pairs.
[Since there is a Ist card and IInd card we have to take ordered pairs]
p(Ist card = IInd card + 1)
= P{(2, 1), (3, 2), (4, 3), (5, 4)} = \(\frac { 4 }{ 5 × 4 }\) = \(\frac { 1 }{ 5 }\)


Question 39.
Consider a random variable X that takes values +1 and -1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = -1 and +1 are (a) 0 and 0.5
(b) 0 and 1
(c) 0.5 and 1
(d) 0.25 and 0.75

Answer/Explanation

Answer:
(c) 0.5 and 1
Explanation:
The p.d.t of the random variable is

Probability Questions and Answers Computer Science Quiz chapter 5 img 14

The cumulative distribution function F(x) is the probability up to x as given below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 15

So the correct option is (c).


Probability Questions and Answers | Computer Science Quiz

Question 40.
Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3 then the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
(a) \(\frac { 10 }{ 21 }\)
(b) \(\frac { 5 }{ 12 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { 1 }{ 6 }\)

Answer/Explanation

Answer:(b) \(\frac { 5 }{ 12 }\)
Explanation:
If first throw is 1, 2 or 3 then sample space is only 18 possible ordered pairs. Out of this only (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5) and (3, 6) i.e. 9 out of 18 ordered pairs gives a Sum ≥ 6.
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:

Probability Questions and Answers Computer Science Quiz chapter 5 img 16

From above diagram
p(sum ≥ 6) = \(\frac { 1 }{ 2 }\) × \(\frac { 9 }{ 18 }\) + \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 3 }\) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 2 }\)


Question 41.
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM, and p has Poisson distribution with a mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
(a) 8/(2e3)
(b) 9/(2e3)
(c) 17/(2e3)
(d) 26/(2e3)

Answer/Explanation

Answer:
(c) 17/(2e3)
Explanation:
Poisson formula for (P = x) given as \(\frac{\mathrm{e}^{-\lambda} \lambda^{\mathrm{x}}}{\mathrm{x} !}\)
α = 3 cars/minute
△T = 1 minute
So λ = α △ T = 3 × 1 = 3
Probability of observing fewer than 3 cars = P(X < 3) = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{\mathrm{e}^{-3} 3^{0}}{0 !}\) + \(\frac{\mathrm{e}^{-3} 3^{1}}{1 !}\) + \(\frac{e^{-3} 3^{2}}{2 !}\) = \(\frac{17}{2 \mathrm{e}^{3}}\)
Option (c) is correct.


Question 42.
Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick
is___________ .

Answer/Explanation

Answer:
Explanation:
Suppose you break a stick of unit length at a point chosen uniformly at random, then let the length of the shorter stick = x
x has uniform distribution in the interval [0,1/2] i.e. a = 0 and p – 1/2
In uniform distribution, E(x) = (β + α)/2
So the expected length of the shorter stick = E(x) = (1/2 + 0)/2 = 1/4 = 0.25


Probability Questions and Answers | Computer Science Quiz

Question 43.
Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is \(\frac { x }{ 1296 }\) The value of x is _________.

Answer/Explanation

Explanation:
Sample space = 64= 1296
6,6,6,4 ⇒ 4 ways (\(\frac { 4 ! }{ 3 !}\) = 4)
6,6,5,5 ⇒ 6 ways (\(\frac { 4 ! }{ 2!2!}\) = 6)
probability of sum to be 22 = \(\frac { 6 + 4 }{ 1296 }\) = \(\frac { x }{ 1296 }\)
⇒ x = 10</de


Question 44.
The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p.
Then 100p = __________.

Answer/Explanation

Explanation:
The tree diagram for the problem is shown below.

Probability Questions and Answers Computer Science Quiz chapter 5 img 17

Required probability = \(\frac{{ }^{4} \mathrm{C}_{3} \cdot{ }^{6} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{4}}\) + \(\frac{{ }^{4} \mathrm{C}_{3} \cdot{ }^{6} \mathrm{C}_{0}}{{ }^{10} \mathrm{C}_{4}}\) = \(\frac { 24 }{ 210}\)
+ \(\frac { 1 }{ 210 }\) = \(\frac { 25 }{ 210 }\)
p = 0.1190
⇒ 100 p = 11.90


Question 45.
The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _________.

Answer/Explanation

Explanation:
1 ≤ x ≤ 100
P(x is not divisible by 2, 3 or 5) = 1 – P(x is divisible by 2, 3 or 5)

Probability Questions and Answers Computer Science Quiz chapter 5 img 18


Question 46.
Let S be a sample space and two mutually exclusive events A and B be such that A ∪ B = S. If P(.) denotes the probability of the event, the maximum value of P(A) P(B) is ___________.

Answer/Explanation

Explanation:
It is given that A and B are mutually exclusive also it is given that A ∪ B = S which means that A and B are collectively exhaustive. Now if two events A and B are both mutually exclusive and collectively exhaustive, then P(A) + P(B) = 1 ⇒P(B) = 1-P(A)
Now we wish to maximize P(A) P(B) = P(A)(1-P(A))
Let, P(A) = x
Now P(A) (1 – P(A)) = x(1 – x) = x – x2
Say y = x- x2
\(\frac { dy }{ dx }\) = 1 – 2x = 0 ⇒ x = \(\frac { 1 }{ 2 }\)
= \(\frac { d^2y }{ dx^2 }\) = -2 < 0; (\(\frac { d^2y }{ dx^2 }\))x = \(\frac { 1 }{ 2 }\) = -2 < 0
y has maximum at x = \(\frac { 1 }{ 2 }\)
ymax = \(\frac { 1 }{ 2 }\) – (\(\frac { 1 }{ 2 }\))2 = 0.25


Question 47.
Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi = 0] = Pr[Xi = 1] = 1/2 for i = 1, 2, 3 Define another random variable Y=X1X2 ⊕ X3, where ⊕ denotes XOR. Then Pr[Y =0 |X3 = 0] =

Answer/Explanation

Explanation:
The p.d.t. for X1, X2, and X3 as given in the problem is shown below:

Probability Questions and Answers Computer Science Quiz chapter 5 img 19

Given, Y = X1X2 ⊕ X3
The required probability – p(Y = 0 | X3 = 0)
= \(\frac{p\left(Y=0 \cap X_{3}=0\right)}{p\left(X_{3}=0\right)}\) …..(1)
Now, p{X3 = 0) = \(\frac { 1 }{ 2 }\) (from p.d.t. of X3) …(2)
p(Y = 0 ∩ X3 = 0)
= p (X1X2 ⊕ X3 = 0 ∩ X3 = 0)
= P((X1X2 ⊕ 0 ) = 0 ∩ X3 = 0)
= p(X1X2 = 0 ∩ X3 = 0)
= p(X1 = 0, X2 = 0, X3 = 0) + p(X1 = 0, X2 = 1, X3 = 0) + p(X1 = 1, X2 = 0, X3 = 0) + P(X1 = 1, X2 = 0, X3 = 0)
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 8 }\)
So, p(Y = 0 ∩ X3 = 0) = \(\frac { 3 }{ 8 }\) ….(3)
Now substituting (2) and (3) in (1), we get
The required probability = p(Y = 0|X3 = 0) = \(\frac { 3/8 }{ 1/2 }\) = \(\frac { 3 }{ 4 }\) = 0.75


Probability Questions and Answers | Computer Science Quiz

Question 48.
A probability density function on the interval [a, 1] is given by 1/x2 and outside this interval, the value of the function is zero. The value of a is ___________.

Answer/Explanation

Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 20


Question 49.
Consider the following experiment:
Step 1. Flip a fair coin twice.
Step 2. If the outcomes are (TAILS, HEADS) then output Yand stop.
Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output Wand stop.
Step 4. If the outcomes are (TAILS, TAILS), then go to Step 1.
The probability that the output of the experiment is Y is __________(up to two decimal places)

Answer/Explanation

Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 21

The tree diagram for the problem is given above. The desired output is Y. Now by rule of total probability P(output = Y) = 0.5 × 0.5 + 0.5 × 0.5 × 0.5 × 0.5 +… Infinite geometric series with α = 0.5 × 0.5 and r = 0.5 × 0.5 So p (output = Y)
= \(\frac { 0.5 × 0.5 }{ 1 -0.5 × 0.5 }\) = \(\frac { 0.25 }{ 0.75 }\) = \(\frac { 1 }{ 3 }\) = 0.33 (upto 2 decimal places)


Question 50.
Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb is chosen uniformly at random lasts more than 100 hours is ___________.

Answer/Explanation

Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 22

P(losts > 100 hr) = 0.5 × 0.7 + 0.5 × 0.4 = 0.35 + 0.2 = 0.55


Question 51.
Let X be a Gaussian random variable with mean 0 and variance σ2. Let Y = max (X, 0) where max (a, b) is the maximum of a and b. The median of Y is ___________.

Answer/Explanation

Explanation:
The Gaussian random variable is the same as the normal random variable.
So, the distribution of X is N(0, σ2)
For X, Median = Mean = Mode = 0 For Y = Max (X, 0)
Now half the data are below 0 and half the data above 0 for X. If we apply Y = Max(X, 0) now, all the negative values will become 0 and all the positive values will remain positive. So, the Median being positional average will not be affected and will remain at 0, since now half will be at 0 and a half will be positive. So, Median (Y) = 0.


Question 52.
P and Q are considering to apply for a job. The probability that P applies for the job is \(\frac { 1 }{ 4 }\) , the probability that P applies for the job given that Q applies for the job is \(\frac { 1 }{ 2 }\) , and the probability that Q applies for the job given that P applies for the job is \(\frac { 1 }{ 3 }\). Then the probability that P does 3 not apply for the job given that Q does not apply for the job is
(a) \(\frac { 4 }{ 5 }\)
(b) \(\frac { 5 }{ 6 }\)
(C) \(\frac { 7 }{ 8 }\)
(d) \(\frac { 11 }{ 12 }\)

Answer/Explanation

Answer:(a) \(\frac { 4 }{ 5 }\)
Explanation:
Given that,
p(P) = \(\frac { 1 }{ 4 }\) …..(1)
p(P | Q) = \(\frac { 1 }{ 2 }\) …..(2)
p(Q | p) = \(\frac { 1 }{ 3 }\) …..(3)
p(\(\bar{P}\) | \(\bar{Q}\) = ?
First solve for p(Q) and p (P ∩ Q) from equation (2) and (3) as follows:
From equation (2)
p(P | Q) = \(\frac { p(P ∩ Q) }{ P(Q) }\) = \(\frac { 1 }{ 2 }\) …(4)
From equation (3)
p(Q | P) = \(\frac { p(P ∩ Q) }{ P(P) }\) = \(\frac { 1 }{ 3 }\)
⇒ p(P ∩ Q) = \(\frac { 1 }{ 3 }\) × p(P) = \(\frac { 1 }{ 3 }\) × \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 12 }\)
Now substitute in equation (4) and get
P(Q) = \(\frac { p(P ∩ Q) }{ 1/2 }\) = \(\frac { 1/12 }{ 1/2 }\) = \(\frac { 2 }{ 12 }\) = \(\frac { 1 }{ 6 }\)
So now we have p(P) = \(\frac { 1 }{ 4 }\)
p(Q) = \(\frac { 1 }{ 6 }\)
and p(P ∩ Q) = \(\frac { 1 }{ 12 }\) we need to find
p(\(\bar{P}\) | \(\bar{Q}\) = \(\frac{p(\bar{P} \cap \bar{Q})}{p(\bar{Q})}\)
= \(\frac{1-(P \cup Q)}{1-p(Q)}\) = 1 – \(\frac{[p(P)+p(Q)-p(P \cap Q)]}{1-p(Q)}\)
= \(\frac{1-\left[\frac{1}{4}+\frac{1}{6}-\frac{1}{12}\right]}{1-\frac{1}{6}}\) = \(\frac{\frac{2}{3}}{\frac{5}{6}}\) = \(\frac { 4 }{ 5 }\)
So, p(\(\bar{P}\) | \(\bar{Q}\) = \(\frac { 4 }{ 5 }\)


Probability Questions and Answers | Computer Science Quiz

Question 53.
For any discrete random variable X, with probability mass function P(X = j) = pi, pj ≥ 0, j∈ {0,…, N}, and \(\sum_{j=0}^{N}\) pj = 1, define the polynomial function gx(z) = \(\sum_{j=0}^{N}\) PjZj For a certain discrete random variable Y, there exists a scalar β ∈ [0,1] such that gy (z) = (1- β + βz)N. The expectation of F is
(a) Nβ (1 – β)
(b) Nβ
(c) N (1 – β)
(d) Not expressible in terms of N and β alone

Answer/Explanation

Answer:(b) Nβ
Explanation:
gy(z) = ((1 – β) + βz)N
If gy(z) is expanded, we would get a bionomial distribution with n = N and p = β So the E(Y) = np = Nβ


Question 54.
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)2] equals __________.

Answer/Explanation

Explanation:
Given, Poisson distribution λ = 5 We know that in Poisson distribution E (X) = V(X) = λ
So here E(X) = V(X) = 5 Now, we need E[(X + 2)2]
= E(X2 +4X + 4)
= E(X2) + 4E(X) + 4
To find E(X2) we write, V(X) = E(X2) – (E(X))2
5 = E(X2) – 52 So, E(X2) = 52 + 5 = 30
Required value = 30+ 4 × 5 + 4 = 54


Question 55.
Two people, P, and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equiprobable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ___________.

Answer/Explanation

Explanation:
P(one of them wins in 3rd trial)
= P(Ist trial is Tie) × P(IInd trial is Tie) × P(one of them wins 3rd trial)
P(Tie in any trial) = P(P = 1 and Q = 1) + P(P = 2 and Q = 2) + …. + P(P = 6 and Q = 6)
= \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) + \(\frac { 1 }{ 36 }\) +\(\frac { 1 }{ 36 }\) = \(\frac { 1 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
P(one of them wins) = 1 = p(Tie) = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
So required probability = \(\frac { 1 }{ 6 }\) × \(\frac { 1 }{ 6 }\) × \(\frac { 5 }{ 6 }\) = \(\frac { 5 }{ 216 }\) = 0.023
(rounded to 3 decimal places)


Probability Questions and Answers | Computer Science Quiz

Question 56.
Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (H), medium (M), and low (L). Let P(HG) denote the probability that Guwahati has a high temperature. Similarly, P(MG) and P(LG) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(HD), P(MD), and P(LD) for Delhi. The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.

HD MD LD
HG 0.40 0.48 0.12
MG 0.10 0.65 0.25
LG 0.01 0.50 0.49

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (HG) then the probability of Delhi also having a high temperature (HD) is 0.40; i.e., P(HD | HG) = 0.40. Similarly, the next two entries are P(MD\HG) = 0.48 and P(LD | HG) = 0.12. Similarly for the other rows.
If it is known that P(HG) = 0.2, P(MG) = 0.5 and P(LG) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is .

Answer/Explanation

Explanation:
The condition probability table given is

HD MD LD
HG 0.40 0.48 0.12
MG 0.10 0.65 0.25
LG 0.01 0.50 0.49

P(HG) = 0.2, P(MG) = 0.5, P(LG) = 0.3 Drawing the tree diagram for HD we get,

Probability Questions and Answers Computer Science Quiz chapter 5 img 23

From diagram, P(HG ∩ HD) = 0.2 × 0.4
P(HD) = 0.2 × 0.4+ 0.5 × 0.1 + 0.3 × 0.01 = 0.133 Required probability,
P(HG | HD) = \(\frac { 0.2 × 0.4 }{ 0.133 }\) = 0.60 (rounding upto 2 decimal place)


Question 57.
Two numbers are chosen independently and uniformly at random from the set {1, 2, …, 13}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is ____________.

Answer/Explanation

Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 24

p(2 elements out of 13 elements chosen randomly and independently) = (Number of MSB‘0’ × Number of MSB‘0’ + Number of MSB‘1’ × Number of MSB‘1’) / (Total × Total) = \(\frac{7 \times 7+6 \times 6}{13 \times 13}\) = \(\frac{49+36}{169}\) = \(\frac{85}{169}\) = 0.503


Question 58.
Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x2 + 6xY+ 3Y+ 6 has only real roots is (rounded off to 1 decimal place) __________.

Answer/Explanation

Explanation:
polynomial 3x2 + 6xY + 3Y + 6 has only real roots
b2 – 4ax ≥ 0
(6Y)2 – 4(3) (3Y + 6) ≥ 0
Y2 – Y + 2 ≥ 0
Y ∈ (-∞, – 1] ∩ [2, ∞)
⇒ Y ∈ [2 , 6)
Since y is uniformly distributed in (1, 6) probability distributed function f(Y) = \(\frac { 1 }{ 5 }\) 1 p(2 ≤ y < 6) = \(\int_{2}^{6}\)f(Y)dy = \(\frac { 1 }{ 5 }\) \([Y]_{2}^{6}\) = \(\frac { 4 }{ 5 }\) = 0.8


Question 59.
For n > 2, let a e {0,1}n be a non-zero vector. Suppose that x is chosen uniformly at random from {0, 1}n. Then, the probability that \(\sum_{i=1}^{n}\)aixian odd number is _________.

Answer/Explanation

Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 25

Fixed non-Zero
ai = 0 or 1
xi = 0 or 1
\(\sum_{i=1}^{n} a_{i} x_{i}\) value lies between 0 to n.
Total number cases = 2n

Aixi 0 1 2 …………… N
fav nC0 nC1 nC2 …………… nCn

Probability Questions and Answers Computer Science Quiz chapter 5 img 26


Question 60.
Consider the two statements:
S1: There exist random variables X and Y such that(E[X-E(X)) (Y-E(Y))]2)>Var[X] Var[Y])
S2 : For all random variables X and Y Cov[X, Y] = E[| X-E[X] || Y-E[Y] |] Which one of the following choices is correct?
(a) Both S1 and S2 are false.
(b) Both S1 and S2 are true.
(c) S1 is true, but S2 is false.
(d) S1 is false, but S2 is true.

Answer/Explanation

Answer:
(a) Both S1 and S2 are false.
Explanation:
S2 : Cov(x,y) = E{|x – \(\bar{x}\)||y – \(\bar{y}\)|} is false
Case-I: If x > \(\bar{x}\) and y > \(\bar{y}\) then above is true.
Case-II: If x < \(\bar{x}\) and y < \(\bar{y}\) then above is true. Case-III: If x > \(\bar{x}\) but y < \(\bar{y}\) then above is false.
Case-IV: If x < \(\bar{x}\) but y > \(\bar{y}\) then above is false.
∵ Given expression is not always true. So we can conclude that it is false.
S1: It is obviously false,
∵ True statement is
[E{(x – \(\bar{x}\))(y – \(\bar{y}\))}]2 < var(x) . Var(y)
So both S1 and S2 are false.


Question 61.
The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter 2. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is __________.

Answer/Explanation

Explanation:
Let, t = {lifetime of component} and μ = 2
Expected lifetime = \(\frac { 1 }{ μ }\)

Probability Questions and Answers Computer Science Quiz chapter 5 img 28
Question 62.
A sender (S) transmits a signal, which can be one of the two kinds: H and L with probabilities 0.1 and 0.9 respectively, to a receiver (R). In the graph below, the weight of edge (u, v) is the probability of receiving v when u is transmitted, where u,v ∈ {H, L}. For example, the probability that the received signal is L given the transmitted signal was H, is 0.7.

Probability Questions and Answers Computer Science Quiz chapter 5 img 29

If the received signal is H, the probability that the transmitted signal was H (rounded to 2 decimal places) is .
Answer:
Explanation:

Probability Questions and Answers Computer Science Quiz chapter 5 img 30


Probability Questions and Answers | Computer Science Quiz

Question 63.
For a given biased coin, the probability that the outcome of a toss is ahead is 0.4. This coin is tossed 1,000 times. Let X denote the random variable whose value is the number of times that head appeared in these 1,000 tosses. The standard deviation of X (rounded to 2 decimal places) is ___________.

Answer/Explanation

Explanation:
n = 1000, p = 0.4, q = 0.6
It is a binomially distributed random variable.
So, S.D. = \(\sqrt{n p q}\)
= \(\sqrt{1000 \times 0.4 \times 0.6}\) = \(\sqrt{240}\) = 15.49


Question 64.
In an examination, a student can choose the order in which two questions (QuesA and QuesB) must be attempted.

  • If the first question is answered wrong, the student gets zero marks.
  • If the first question is answered correctly and the second question is not answered correctly, the student gets the marks only for the first question.
  • If both the questions are answered correctly, the student gets the sum of the marks of the two questions.

The following table shows the probability of correctly answering a question and the marks of the question respectively.

Question Probability of answering correctly Marks
QuesA 0.8 10
QuesB 0.5 20

Assuming that the student always wants to maximize her expected marks in the examination, in which order should she attempt the questions, and what are the expected marks for that order (assume that the questions are independent)?
(a) First QuesA and then QuesB. Expected marks 14.
(b) First QuesB and then QuesA. Expected marks 22.
(c) First QuesB and then QuesA. Expected marks 14.
(d) First QuesA and then QuesB. Expected marks 16.

Answer/Explanation

Answer:
(d) First QuesA and then QuesB. Expected marks 16.
Explanation:
X → Random variable which represents total marks record.
P(X) → Probability of getting those marks.

X → 0 10 20 30
P(x) → 0.2 × 0.5 0.8 × 0.5 0.5 × 0.2 0.8 × 0.5
11 11 11 11
0.1 0.4 0.1 0.4
ΣP(x) = 1

Now, if QuestionA is attempted first and it is correct.
Case-I:
E(x) = Σ(x) . P(x)
= 0.4 × 10 + 0.4 × 30 = 4 + 12 = 16
Case-II:
If QuestionB is attempted first and is correct.
E(x) – Σ(x) . P(x)
= 0.1 (20) + 0.4 (30)
= 2 + 12 = 14

So case-I is giving maximum expected marks. Hence option (d) is correct.


Question 65.
A bag has r red balls and b black balls. All balls are identical except for their colors. In a trial, a ball is randomly drawn from the bag, its color is noted and the ball is placed back into the bag along with another ball of the same color. Note that the number of balls in the bag will increase by one, after the trial. A sequence of four such trials is conducted. Which one of the following choices gives the probability of drawing a red ball in the fourth trial?

Probability Questions and Answers Computer Science Quiz chapter 5 img 32

Answer/Explanation

Answer:
Explanation:
There are 10 favorable ways to calculate the probability of red ball in 4th trial.
(RFR)R = R or (BRR)R ≈ 1 way or (RRR)R ≈ 3 ways or (BBR)R ≈ 3 ways

Probability Questions and Answers Computer Science Quiz chapter 5 img 31


Computer Science Combinatorics Questions and Answers

Combinatorics Questions with Answers for Computer Science

Computer Science Combinatorics Questions and Answers

Question 1.
(a) Solve the recurrence equations:
T(n) = T(n – 1) + n
T(1) = 1
(b) What is the generating function G(z) for the sequence of Fibonacci numbers?

Answer/Explanation

Explanation:

(a) T(n) = T(n – 1) + n

T2 – Tn – 1 n
For Homogeneous solution:
Tn -Tn – 1 = 0
t- 1 = 0
t = 1
Therefore, homogenous solution is
Tn = C(1)n = C
For Particular solution:
Let particular solution be (d0 + d1n)n
⇒ (d0 + d1n)n – (d0 + d1n – 1))(n – 1) = n
⇒ d0n + d1n2 – d0n + d0 – d1(n – 1)2 = n
⇒ d0n + d1n2 – d0n + d0 – d1(n2– 2n + 1) = n
⇒ d0 + 2d1n – d1 = n
d0 – d1 = 0 and 2d1 = 1
⇒ d0 = d1 and d1 = \(\frac{1}{2}\)
∴ d0 = d1 = \(\frac{1}{2}\)
So particular solution is,
\(\left(\frac{1}{2}+\frac{1}{2} n\right)\) × n = \(\frac{n(n+1)}{2}\) = \(\frac{n^{2}+n}{2}\)
So complete solution is,
T(n) C +\(\frac{1}{2}\)n{n +1)
Given, T( 1) = 1
1 = C + \(\frac{1}{2}\) × 1 × (1 + 1)
1 = C+ 1
C= 0
Therefore complete solution of the recurrence relation is T(n) = \(\frac{n(n+1)}{2}\).

(b) The Fibonacci numbers are defined as a0 = 1, a1 = 1
ar = ar – 1 + ar – 2 r ≥ 2

Computer Science Combinatorics Questions and Answers chapter 3 img 1

⇒ A(x) – a0 – a1 x = (A(x) – a0) + x2A(x)
Since, a0 = 1 and a1 = 1
⇒ A(x) – 1 – x = x(A(x) – 1) + x2A(x)
⇒ A(x) = \(\frac{1}{1-x-x^{2}}\)


Question 2.
Solve the recurrence equations:
T(n) = T\(\left(\frac{\mathrm{n}}{2}\right)\) + 1
T(1) = 1

Answer/Explanation

Explanation:
T(n) = T\(\left(\frac{\mathrm{n}}{2}\right)\) + 1
⇒ T(n) – T\(\left(\frac{\mathrm{n}}{2}\right)\) = 1
Let, n = 2k.
⇒ T(2k) – T\(\left(\frac{2^{\mathrm{k}}}{2}\right)\) = 1
⇒ T(2k) – T(2k – 1 = 1
Let, T(2k) – xk
xk – xk – 1 = 1
For Homogeneous solution:
xk – xk – 1 = 0
t – 1 = 0
t = 1
So homogeneous solution is: xk = C(1)k = C
For Particular solution:
Let particular solution be d1k
d1k – d1(k- 1) = 1
⇒ d1 = 1
Particular solution is k
∴ Complete solution is
xk = C + k
T(2k) = C + k
T(n) = C + log2n
Given, T(1) = 1
⇒ C= 1
∴ Complete solution is: T(n) = log2n + 1


Question 3.
How many substrings can be formed from a character string of length n?

Answer/Explanation

Explanation:
Let the string be of length 4 : abcd
Number of substrings of length 0=1 (only ε )
Number of substrings of length 1 = 4
a, b, c, d
Number of substrings of length 2 = 3
ab, bc, cd
Number of substrings of length 3 = 2
abc, bcd
Number of substrings of length 4=1
abcd
∴ Total number of substrings = 1 +(4+ 3 + 2 + 1)
= 1 + (Sum of 4 natural number)
= 1 + \(\frac{4 \times(4+1)}{2}\)
= 11
Therefore, total number of substrings (maximum) that can be formed from a character string of
length 1 + \(\frac{n(n+1)}{2}\).


Computer Science Combinatorics Questions and Answers

Question 4.
The number of binary strings of n zeros and k ones that no two ones are adjacent is
(a) n+1CK
(b) nCK
(c) nCK+1
(d) None of these

Answer/Explanation

Answer: (a) n+1CK
Explanation:
First arranging all n zeros in a row. There is only 1 way for arranging n zeros in a row. By arranging n zeros in a row, we get (n + 1) positions to place ones.
So number of ways arranging k ones in (n + 1) positions = n + 1Ck
∴ Required number of binary strings of n zeroes and k ones that no two ones are adjacent = 1 × n + 1Ck = n + 1Ck.


Question 5.
The number of substrings (of all lengths inclusive) that can be formed from a character string of length n is
(a) n
(b) n2
(c) \(\frac{n(n-1)}{2}\)
(d) \(\frac{n(n+1)}{2}\)

Answer/Explanation

Answer: (d) \(\frac{n(n+1)}{2}\)
Explanation:
For a string of length n:
The number of substrings of length 1 = n
The number of substrings of length 2 = n – 1
The number of substrings of length 3 = n – 2 and so on…
The number of substrings of length n is 1 So total number of substrings ; = n + (n-1) + …+1
= Sum of n natural numbers.
= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)


Question 6.
In a room containing 28 people, there are 18 people who speak English, 15 people who speak Hindi and 22 people who speak Kannada. 9 persons speak both English and Hindi, 11 people speak both Hindi and Kannada whereas 13 persons speak both Kannada and English. How many people speak all three languages?
(a) 9
(b) 8
(c) 7
(d) 6

Answer/Explanation

Answer: (d) 6
Explanation:
E : Persons speaks English
H : Persons speaks Hindi
K : Persons speaks Kannada
Assume that everyone in the room speaks at least one of the languages.
Given, n(E ∪ H ∪ K) = 28
n(E) = 18
n(H) =15
n(K) = 22
n(E ∩ H) = 9
n(H ∩ K) = 11
n(K ∩ E) = 13
n(K ∩ E ∩ H) = ?
By principle of inclusion and exclusion
n(E ∪ H ∪ K) = n(E) + n(H) + n(K) – n(E ∩ H) – n(H ∩ E) – n(K ∩ E) + n(K ∩ E ∩ H)
28 = 18 + 15 + 22 – 9 – 11 – 13 + n(K ∩ E ∩ H) n(k ∩ E ∩ H) = 28 – 55 + 33 = 61 – 55 = 6
Therefore, the number of people who speak all three languages are 6.


Computer Science Combinatorics Questions and Answers

Question 7.
Solve the following recurrence relation
xn = 2xn-1, n>1
x1 = 2

Answer/Explanation

Explanation:
xn = 2xn – 1 – 1
For Homogeneous solution:
xn = 2xn – 1 = 0
t – 2 = 0
t= 2
So homogeneous solution is
xn = C(2)n
For Particular solution:
Let particular solution be do.
d0 = 2d0 – 1
d0 = 1
So particular solution is 1
∴ Complete solution = Homogeneous solution + Particular solution
Complete solution = C(2)n + 1
⇒ xn = C(2)n + 1
Given initial condition is x1 = 2
2 = C(2)1 + 1
1 = 2C
C = \(\frac { 1 }{ 2 }\)
xn = \(\frac { 1 }{ 2 }\)(2)n + 1
∴ xn = 2n – 1 + 1


Question 8.
Two girls have picked 10 roses, 15 sunflowers and 14 daffodils. What is the number of ways they can divide the flowers among themselves?
(a) 1638
(b) 2100
(c) 2640
(d) None of these

Answer/Explanation

Answer: (c) 2640
Explanation:
Number of ways for distributing r similar things among n different things = (n – 1 + r)Cr
The number of ways for distributing 10 roses among the two girls = (2 – 1 + 10)C10 =11.
Similarly number of ways for distributing 15 sunflowers among two girls = (2 – 1 + 15)C15
= 16C15 = 16C1 = 16
Number of ways for distributing 14 daffodils among the two girls = (2 – 1 + 14)C14 = 15C14 = 15C1 =15
∴ Total number of ways = 11 × 16 × 15 = 2640


Question 9.
The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from some same suit is
(a) 3
(b) 8
(c) 9
(d) 12

Answer/Explanation

Answer: (c) 9
Explanation:
Let the number of cards to be dealt from an arbitrarily shuffled deck of 52 cards be n.
Number of suits = 4
Required number of cards from the same suit = 3.
So by the pigeonhole principle

Computer Science Combinatorics Questions and Answers chapter 3 img 2

So the minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from the same suit is 9.


Computer Science Combinatorics Questions and Answers

Question 10.
How many 4 digit even numbers have all 4 digits distinct?
(a) 2240
(b) 2296
(c) 2620
(d) 4536

Answer/Explanation

Answer: (b) 2296
Explanation:
The digits are given to be distinct i.e. no repetition. 4 digit even numbers cannot start with 0 and must end with 0, 2, 4, 6 or 8.
Since there is a condition for 0 in starting as well as ending we will count the even numbers ending with 0 separately.
So the total number of 4 digits even number = 4 digit even numbers ending with zero + 4 digit even numbers ending with 2, 4, 6, 8.
4 digit even numbers ending with 0

Computer Science Combinatorics Questions and Answers chapter 3 img 3

4 digit even numbers ending with 2, 4, 6, 8

Computer Science Combinatorics Questions and Answers chapter 3 img 4

So the total number of 4 digit even numbers = 504+1792 = 2296.


Question 11.
The solution to the recurrence equation T(2k) = 3T(2k-1) + 1, T(1) = 1 is
(a) 2k
(b) \(\frac{\left(3^{\mathrm{k}+1}-1\right)}{2}\)
(c) 3log2k
(d) 2log3k

Answer/Explanation

Answer: (b) \(\frac{\left(3^{\mathrm{k}+1}-1\right)}{2}\)
Explanation:
T(2k) = 3T(2k – 1) + 1
Let, T(2k) = xn
⇒ xn = 3xn – 1 + 1
⇒ xn – 3xn – 1 = 1
So for Homogenous solution
xn – 3xn – 1 = 0
n – 3 = 0
n = 3
Homogenous solution is
xn = C1(3)n
T(2k) = C1(3)k
For Particular solution
Let d be the particular solution
d – 3d = 1
2d = -1
d = \(-\frac{1}{2}\)
Therefore, the complete solution is
T(2k) = C1(3)k \(-\frac{1}{2}\)
Given T(1) = 1
1 = C1(3)0 \(-\frac{1}{2}\)
1 = C1 \(-\frac{1}{2}\)
C1 = \(\frac { 3 }{ 2 }\)
So the complete solution is
T(2k) = C1 (3)k \(-\frac{1}{2}\)
Given, T(1) = 1
1 = C1(3)0 \(-\frac{1}{2}\)
1 = C1 \(-\frac{1}{2}\)
C1 = \(\frac { 3 }{ 2 }\)
So the complete solution is
T(2k) = \(\frac { 3 }{ 2 }\)(3)k \(-\frac{1}{2}\)
T(2k) = \(\frac{3^{\mathrm{k}+1}-1}{2}\)


Question 12.
The minimum number of colors required to color the vertices of a cycle with n nodes in such a way that no two adjacent nodes have the same color is
(a) 2
(b) 3
(c) 4
(d) n – 2\(\left[\frac{n}{2}\right\rfloor\)+2

Answer/Explanation

Answer: (d) n – 2\(\left[\frac{n}{2}\right\rfloor\)+2
Explanation:
The minimum number of colors required to color the vertices of a cycle with n nodes = 2, when n is even = 3, when n is odd
Therefore n – 2 \(\left\lfloor\frac{\mathrm{n}}{2}\right\rfloor\) +2 gives 2 when n is even. and 3 when n is odd.


Computer Science Combinatorics Questions and Answers

Question 13.
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that (i) each are sorted in ascending order, (ii) B has 5 and C has 3 elements, and (iii) the result of merging B and C gives A?
(a) 2
(b) 30
(c) 56
(d) 256

Answer/Explanation

Answer: (c) 56
Explanation:
This corresponds to an ordered partition of 8 elements into two groups, the first with 5 elements and second with 3 elements. The number of ways of doing this is p(8; 5, 3) = \(\frac{8 !}{5 ! 3 !}\) = 56


Question 14.
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is
(a) \(\left(\begin{array}{c}
2 n \\
n
\end{array}\right)\) * 2n
(b) 3n
(c) \(\frac{(2 n) !}{2^{n}}\)
(d) \(\left(\begin{array}{c}
2 n \\
n
\end{array}\right)\)

Answer/Explanation

Answer: (b) 3n
Explanation:
For each of the n couples invited to the party one of three things is possible:
1. Both husband and wife attend the party.
2. Wife only attends the party.
3. Neither husband nor wife attends the party. Since there are n such couples, a total number of possibilities = 3n.


Question 15.
Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour-pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?
(a) 9
(b) 8
(c) 7
(d) 6

Answer/Explanation

Answer: (d) 6
Explanation:
The problem reduces to finding how many distinct ordered color pairs (C1, C2) are possible with k colors.
Since the first color C1 can be any one of the k colors and the second color C2 also can be any one of the k colors (both prints of a letter can be colored with the same color), the total no. of such order color pairs is equal to k × k = k2.
Since each pair of letters must be colored with different color pairs, at least 26 color pairs are required to do this.
Therefore the requirement is k2 ≥ 26.
The minimum value of k that satisfies this equation is k = 6.


Computer Science Combinatorics Questions and Answers

Question 16.
In how many ways can we distribute 5 distinct balls, B1, B2……., B5 in 5 distinct cells, C1 C2, …, C5 such that Ball Bi is not in cell Ci,∀i = 1,2,…,5 and each cell contains exactly one ball?
(a) 44
(b) 96
(c) 120
(d) 3125

Answer/Explanation

Answer: (a) 44
Explanation:
We want every one of the 5 balls to be in the wrong box. This is nothing but the number of derangements of a set of 5 elements = D5. i.e. we need to compute D5

Computer Science Combinatorics Questions and Answers chapter 3 img 5


Question 17.
Let n = p2q, where p and q are distinct prime numbers. How many numbers m satisfy 1 ≤ m ≤ n and gcd (m, n) = 1? Note that gcd (m, n) is the greatest common divisor of m and n.
(a) p(q – 1)
(b) pq
(c) (p2 – 1)(q – 1)
(d) p(p – 1)(q – 1)

Answer/Explanation

Answer: (d) p(p – 1)(q – 1)
Explanation:
The number of numbers from 1 to n, which are relatively prime to n i.e., gcd (m, n) = 1, is given by the Euler Totient function Φ(n). If n is broken down into its prime factors as n = p1n1 . p2n2 ……
where p1,p2 etc. are distinct prime numbers, then

Φ(n) = Φ( p1n1) Φ(p2n2)
Here, n = p2 q
So, Φ(n) = Φ(p2) × Φ(q)
Now, using the property Φ(pk) = pk – pk – 1
Φ(p2) = p2 – p1 = p2 – p and
Φ(q) = q1 – q0 = q – 1
Substituting these in eq. (i), we get
Φ(n) = (p2 – p) (q – 1)
= p(p – 1) (q – 1)


Question 18.
What is the minimum number of ordered pairs of non-negative numbers that should be chosen to ensure that there are two pairs (a, b) and (c, d) in the chosen set such that a=c mod 3 and b = d mod 5
(a) 4
(b) 6
(c) 16
(d) 24

Answer/Explanation

Answer: (c) 16
Explanation:
The number of combinations of pairs (a mod 3, b mod 5) is 3 × 5 = 15 (since a mod 3 can be 0, 1, or 2) and b mod 5 can be 0, 1, 2, 3 or 4)
∴ If 16 different ordered pairs are chosen at least 2 of them must have (a mod 3, b mod 5) as same (basic pigeon hole principle).
Let such two pairs be (a, b) and (c, d) then
a mod 3 ≡ c mod 3 ⇒ a ≡ c mod 3, and b mod 5 ≡ d mod 5 ⇒ b ≡ d mod 5


Question 19.
Let G(x) = 1/(1 – x)2 = \(\sum_{x=1}^{∞}\)g(i)xi where |x| < 1. what is g(i)?
(a) i
(b) i + 1
(c) 2i
(d) 2i

Answer/Explanation

Answer: (b) i + 1
Explanation:

Computer Science Combinatorics Questions and Answers chapter 3 img 6


Computer Science Combinatorics Questions and Answers

Question 20.
For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tossed are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is
(a) \(\left(\begin{array}{c}
2 n \\
\mathrm{n}
\end{array}\right) / 4^{\mathrm{n}}\)
(b) \(\left(\begin{array}{c}
2 n \\
n
\end{array}\right) / 2^{n}\)
(c) \(1 /\left(\begin{array}{c}
2 \mathrm{n} \\
\mathrm{n}
\end{array}\right)\)
(d) \(\frac { 1 }{ 2 }\)

Common Data for Q.21 & Q.22
Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i, j) then it can move to either (i + 1, j) or (i, j + 1).

Answer/Explanation

Answer:
(a) \(\left(\begin{array}{c}
2 n \\
\mathrm{n}
\end{array}\right) / 4^{\mathrm{n}}\)
Explanation:
The probability that exactly n elements are chosen = The probability of getting n heads out of 2n tosses
= 2nCn (1/2)n (1/2)2n – n(Binomial formula)
= 2nCn (1/2)n (1/2)n
= \(\frac{{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}}{2^{2 \mathrm{n}}}\) = \(\frac{{ }^{2 n} \mathrm{C}_{\mathrm{n}}}{\left(2^{2}\right)^{\mathrm{n}}}\) = \(\frac{{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}}{4^{\mathrm{n}}}\)


Question 21.
How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0, 0)?
(a) \(\left(\begin{array}{l}
20 \\
10
\end{array}\right)\)
(b)220
(c)210
(d) None of these

Answer/Explanation

Answer:
(a) \(\left(\begin{array}{l}
20 \\
10
\end{array}\right)\)
Explanation:
Consider the following diagram:

Computer Science Combinatorics Questions and Answers chapter 3 img 7

The robot can move only right or up as defined in the problem. Let us denote the right move by ‘R’ and the up move by ‘U’. Now to reach (3, 3) from (0, 0), the robot has to make exactly 3 ‘R’ moves and 3 ‘U’ moves in any order. Similarly to reach (10, 10) from (0, 0), the robot has to make 10 ‘R’ moves and 10 ‘U’ moves in any order. The number of ways this can be done is the same as the number of permutations of a word consisting of 10 ‘R’s and 10 ‘U’s.
Applying formula of permutation with limited repetitions we get the answer as \(\frac{20 !}{10 ! 10 !}\) = 20C10.


Question 22.
Suppose that the robot is not allowed to traverse the line segment from (4, 4) to (5, 4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)?
(a) 29
(b) 219
(c) \(\left(\begin{array}{l}
8 \\
4
\end{array}\right) \times\left(\begin{array}{c}
11 \\
5
\end{array}\right)\)
(d) \(\left(\begin{array}{l}
20 \\
10
\end{array}\right)-\left(\begin{array}{l}
8 \\
4
\end{array}\right) \times\left(\begin{array}{c}
11 \\
5
\end{array}\right)\)

Answer/Explanation

Answer:
Explanation:
The robot can reach (4, 4) from (0,0) in 8C4 ways as argued in the previous problem. Now after reaching (4,4), the robot is not allowed to go to (5,4). Let us count how many paths are there from (0,0) to (10, 10) if the robot goes from (4, 4) to (5, 4) and then we can subtract this from a total number of ways to get the answer.
Now there are 8C4 ways for robot to reach (4, 4) from (0, 0) and then robot takes the ‘U’ move from (4, 4) to (5, 4). Now from (5, 4) to (10, 10) the robot has to make 5 ‘U’ moves and 6 ‘R’ moves
in any order which can be done in \(\frac{11 !}{5 ! 6 !}\) ways = 11C5 ways.

∴ The number of ways robot can move from (0,0) (10, 10) via (4, 4) – (5, 4) move is
8C4 × 11C5 = \(\left(\begin{array}{l}
8 \\
4
\end{array}\right)\) × \(\left(\begin{array}{c}
11 \\
5
\end{array}\right)\)

∴ Number of ways robot can move from (0,0) to (10, 10) without using (4, 4) to (5, 4) move is

\(\left(\begin{array}{l}
20 \\
10
\end{array}\right)\) – \(\left(\begin{array}{l}
8 \\
4
\end{array}\right)\) x \(\left(\begin{array}{l}
11 \\
5
\end{array}\right)\) ways.
Which is an option (d).


Computer Science Combinatorics Questions and Answers

Question 23.

Computer Science Combinatorics Questions and Answers chapter 3 img 8

where k is positive integer. Then
(a)P = Q – k
(b)P=Q + k
(c) P = Q
(d) P = Q + 2k

Answer/Explanation

Answer:
Common Data for Q.24 & Q.25
Let xn denote the number of binary strings of length n that contain no consecutive 0s.
Explanation:
P = Σ i = 1 + 3 + 5 + 7 + … + (2k – 1)
1 ≤ i ≤ 2k, i is odd
Q= Σ i=2+4+ 6+…2k, 1 < i < 2k, 1 ≤ i ≤ 2k, i is even.
P is in A.P with a = 1, d = 2 and n = k
Q is in A.P with a = 2, d = 2 and n = k

Computer Science Combinatorics Questions and Answers chapter 3 img 9


Question 24.
Which of the following recurrences does xn satisfy?
(a) xn = 2xn – 1
(b) xn = x[n/2] + 1
(c) xn = x[n/2] + n
(d) xn = xn – 1 + xn – 2

Answer/Explanation

Answer: (d) xn = xn – 1 + xn – 2
Explanation:
The ✓ represents those strings with no consecutive 0’s.

Computer Science Combinatorics Questions and Answers chapter 3 img 10

Now, substituting n = 3 in all of the answers only choice (d) xn = xn – 1 + xn – 2 satisfies the numbers obtained from the tree counting.


Question 25.
The value of x5 is
(a) 5
(b) 7
(c) 8
(d) 13

Answer/Explanation

Answer: (d) 13
Explanation:
x1 = 2
x2 = 3
⇒ x3 = x1 + x2 = 2 + 3 = 5
⇒ x4 = 2 + x3 = 3 + 5 = 8
⇒ x5 = 3 + x4 = 5 + 8 = 13
So option (d) is correct.


Question 26.
The exponent of 11 in the prime factorization of 300! is
(a) 27
(b) 28
(c) 29
(d) 30

Answer/Explanation

Answer: (c) 29
Explanation:
In 300!, distinct numbers divisible by 11 = \(\left[\frac{300}{11}\right\rfloor\) = 27.
Since these 27 numbers contain, the same number which can be further divisible by 11 i.e.
\(\left[\frac{27}{11}\right\rfloor\) = 2(121 and 242)
So, total number of 11’s = 27 + 2 = 29 Hence exponent of 11 is 29.


Question 27.
In how many ways can b blue balls and r red balls be distributed in re distinct boxes?
(a) \(\frac{(n+b-1) !(n+r-1) !}{(n-1) ! b !(n-1) ! r !}\)
(b) \(\frac{(n+(b+r)-1) !}{(n-1) !(n-1) !(b+r) !}\)
(c) \(\frac{n !}{b ! r !}\)
(d) (n + (b + r) -1)! /n!(b + r – 1)

Answer/Explanation

Answer: (a) \(\frac{(n+b-1) !(n+r-1) !}{(n-1) ! b !(n-1) ! r !}\)
Explanation:
A number of ways to distribute ‘b’ blue balls in n distinct boxes:

Computer Science Combinatorics Questions and Answers chapter 3 img 11

Since each box can contain 0 to n blue balls, so number of ways
= ( n + b – 1)C(n – 1)
=\(\frac{(n+b-1) !}{(n-1) ! \times b !}\)

Number of ways to distribute ‘r’ red balls in n distinct boxes:

Computer Science Combinatorics Questions and Answers chapter 3 img 12

Since each box can contain 0 to n red balls. So number of ways = ( n + r – 1)C(n – 1)

=\(\frac{(n+r-1) !}{(n-1) ! \times r !}\)

Since both are independent of each other So total ways = Number of ways to distribute ‘b’ blue balls x Number of ways to distribute ‘r’ red balls

= \(\frac{(n+b-1) !(n+r-1) !}{(n-1) ! \times b !(n-1) ! \times r !}\)


Computer Science Combinatorics Questions and Answers

Question 28.
A pennant is a sequence of numbers, each number being 1 or 2. An n-pennant is a sequence of numbers with a sum equal to n. For example, (1,1,2) is a 4-pennants. The set of all possible 1-pennant is {(1)}, the set of all possible 2-pennants is {(2), (1,1)}and the set of all 3-pennants is {(2,1), (1,1,1), (1,2)}. Note that the pennant (1,2) is not the same as the pennant (2,1). The number of 10-pennants is ______.

Answer/Explanation

Explanation:
In a 10-pennant let there we x1 ones and x2 twos. So we need to find all the solutions of
x1 + 2x2 = 10

put, x1 = 0 ⇒ x2 = \(\frac { 10 }{ 2 }\) = 5

So, (0, 5) is a solution i.e. a 10-pennant could have 0 ones and 5 twos. The number of ordered permutations of 0 ones and 5 twos = 5!/ 5! = 1

Now x1 cannot be 1 since in that case x2 = \(\frac { 10 }{ 2 }\) = 4.5
(is not an integer).

put, x1 = 2 ⇒ x2 = \(\frac { 8 }{ 2 }\) = 4

So, (2, 4) is a solution i.e. a 10-pennant could have 2 ones and 4 twos. The number of ordered permutations of 2 ones and 4 twos

\(\frac{6 !}{2 ! 4 !}\) = 15

Similarly (4, 3), (6, 2), (8, 1) and (10,0) are the other four solutions and the number of pennants for each is respectively

\(\frac{7 !}{4 ! 3 !}\) = 35, \(\frac{8 !}{6 ! 2 !}\) =28, \(\frac{9 !}{8 ! 1 !}\) = 9, \(\frac{10 !}{10 ! 10 !}\) = 1 = 89

So the total number of 10-pennants = 1 + 15 + 35 + 28 + 9 + 1 = 89


Question 29.
Each of the nine words in the sentence
“The quick brown fox jumps over the lazy dog”
is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____. (The answer should be rounded to one decimal place).

Answer/Explanation

Explanation:
“The quick brown fox jumps over the lazy dog” (3) (5) (5) (3) (5) (4) (3) (4) (3)
Now let x be the number of letters in the word that is randomly picked.
Now, we make a probability distribution table for x

Computer Science Combinatorics Questions and Answers chapter 3 img 13

From this table we can easily find the expected value of x.

E(x) = Σx p(x)
= 3 × \(\frac { 4 }{ 9 }\) + 4 × \(\frac { 2 }{ 9 }\) + 5 × \(\frac { 3 }{ 9 }\) = \(\frac { 35 }{ 9 }\) = 3.88 = 3.9
(after rounding to one decimal accuracy).


Question 30.
The number of distinct positive integral factors of 2014 is _______.

Answer/Explanation

Explanation:
Factorizing 2014 in primes by successively dividing by primes, we get
2014 = 21 × 191 × 531
Now We use the formula:
If n = p1n1 . p2n2…..prnr is the prime factorization of n, then the number of distit factors of n is given by (n1 + 1) × (n2 + 1) …….. (nr + 1).
Now since 2014 = 21 × 191 × 531, the number of distinct factors of 2014 = (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 = 8


Question 31.
\(\sum_{x=1}^{99} \frac{1}{x(x+1)}\) = _______.

Answer/Explanation

Explanation:

Computer Science Combinatorics Questions and Answers chapter 3 img 14

= 1 – \(\frac { 1 }{ 100 }\) (all the terms in above series except the first and last terms cancels out)
= \(\frac { 99 }{ 100 }\) = 0.99


Computer Science Combinatorics Questions and Answers

Question 32.
Let an represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for an?
(a) an -2 + an – 1 + 2n – 2
(b) an -2 + 2an – 1 + 2n – 2
(c) 2an -2 + an – 1 + 2n – 2
(d) 2an -2 + 2an – 1 + 2n – 2

Answer/Explanation

Answer: (a) an -2 + an – 1 + 2n – 2
Explanation:
a1 = 0 [n0 strings of length 1 contain two consecutive 1’s]
a2 = 1 [∴ strings = 11]
a3 = 3 [∴ strings are : 011, 110, 111]
a4 = 8 [∴ strings are : 0011, 0110, 0111, 1011, 1100, 1101, 1110, 1111]

Option (a):
an = an – 2 + an – 1 + 2n – 2
⇒ a4 = a4 – 2 + a4 – 1 + 24 – 2
= a2 + a3 + 22
= 1 + 3 + 4 = 8 which is true.

Option (b):
an = an – 2 + 2an – 1 + 2n – 2
⇒ a4 = a2 + 2a3 + 22
= 1 + 2 × 3 + 4 = 11 which is False.

Option (c):
an = 2an – 2 + an – 1 + 2n – 2
⇒ a4 = 2a2 + a3 + 22
= 2 × 1 + 3 + 4 = 9 which is false.

Option (d):
an = 2an – 2 + 2an – 1 + 2n – 2
⇒ a4 = 2a2 + 2a3 + 22
= 2 × 1 + 2 × 3 + 4 = 12 which is false.

∴ Option (a) : an = an – 2 + an – 1 + 2n – 2 is correct.


Question 33.
The number of divisors of 2100 is _______.

Answer/Explanation

Explanation:
Dividing 2100 successively by prime numbers, we get that, the prime factors of 2100 are 2 × 2 × 3 × 5 × 5 × 7
⇒ 22 × 31 × 52 × 71
If N = pk1 × qk2 × rk3…… × ykn
then the number of factors of N are (k1+1) (k2+1) … (kn+1)
∴ For 2100 we have (2 + 1) (1 + 1) (2 + 1) (1 + 1) = 36 factors.


Question 34 .
The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is ____.

Answer/Explanation

Explanation:
To satisfy the non-decreasing order condition allow 1, 2, 3 after 1, allow only 2, 3 after 2 and allow only 3 after 3. The following tree diagram gives all the allowed numbers satisfying the given condition.

Computer Science Combinatorics Questions and Answers chapter 3 img 15

All the tick-marked numbers satisfy the non-decreasing order condition.
⇒ A number of 4-digit numbers = Number of tick marks = 15.


Question 35.
Let an be the number of re-bit strings that do NOT contain two consecutive l’s. Which one of the following is the recurrence relation for an?
(a) an = an – 1 + 2an – 2
(b) an = an – 1 + an – 2
(c) an = 2an – 1 + an – 2
(d) an = 2an – 1 + 2an – 2

Answer/Explanation

Answer: (d) an = 2an – 1 + 2an – 2
Explanation:
Let an be the number of n-bit strings that do not contain two consecutive 1’s. we wish to develop a recurrence relation for an.
Consider 1 bit strings 0, 1 So, a1 = 2
Consider 2 bit strings 00, 01, 10, 11
Out of minimum only 00, 01, 10 do not contain two consecutive 1’s.
So, a2 = 3

Computer Science Combinatorics Questions and Answers chapter 3 img 16

Out of minimum six strings only 000, 001, 010, 100 and 101 five strings satisfy do not contain two consecutive 1’s. So, a3 = 5. Three numbers a1, a2, a3 satisfy clearly only (b) an = an – 1 + an – 2 is correct.


Question 36.
The coefficient of x12 in (x3 + x4 + x5 + x6 +…)3 is _______.

Answer/Explanation

Explanation:
We wish to find coefficient of x12 in (x3 + x4 + x5 +…)3

Computer Science Combinatorics Questions and Answers chapter 3 img 17

Now to make x12 we need to put r = 3 So coefficient of x12 is 3 + 2C3 = 5C3 = 5C2 = 10


Question 37.
Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an – 1 Let a99 = K × 104. The value of K is ______.

Answer/Explanation

Explanation:
Given, an = 6n2 + 2 n + an – 1 and a1 = 8
We wish to find a99
Now,
a2 = 6 × 22 + 2 × 2 + a1
a3 = 6 × 32 + 2 × 3 + a2
= 6 × 32 + 2 × 3 + 6 × 22 + 2 × 2 + a1………
a99 = 6 × 992 + 2 × 99 + 6 × 982 + 2 + 98 ….
…. + 6 × 22 + 2 × 2 + a1
Since,
a1 = 8
a99 = 6 × 992 + 2 × 99 + 6 × 982 + 2 × 98…….
……+ 6 × 22 + 2 × 2 + 8
= 6 × 992 + 2 × 99 + 6 × 982 + 2 × 98…
…6 × 22 + 2 × 2 + 6 × 12 + 2 × 1
= 6(12+22+32… 992)+ 2.(1 + 2 + 3…99)
= 6. \(\frac{(99(99+1)(2 \times 99+1))}{6}\) + 2 \(\left(\frac{99(99+1)}{2}\right)\)
= 99 × 100 × 199 + 99 × 100
= 100 × 99 (199 + 1) = 100 × 99 × 200
= 2 × 99 × 104 = 198 × 104
So if a99 = K × 104 then K = 198.


Question 38.
If the ordinary generating function of a sequence \(\left\{a_{n}\right\}_{n=0}^{\infty} \text { is } \frac{1+z}{(1-z)^{3}}\) , then a3 – a0 is equal to _______.

Answer/Explanation

Explanation:
Given that generating function:

Computer Science Combinatorics Questions and Answers chapter 3 img 18

Now we read to find a0 and 3 which are nothing but the coefficient of x0 and x3 respectively.
a0 = coefficient x0 = (0 + 2)C2 = 2C2 = 1
a3 = coefficient x3 = (3 + 2)C2 = (2 + 2)C2
= 5C2 + 4C2 = 16
So, a3 – a0 = 16 – 1 = 15


Computer Science Combinatorics Questions and Answers

Question 39.
Which one of the following is a closed form expression for the generating function of the sequence {a1n}, where an = 2n + 3 for all n = 0 1,2,…..?
(a) \(\frac{3}{(1-x)^{2}}\)
(b) \(\frac{3 x}{(1-x)^{2}}\)
(c) \(\frac{2-x}{(1-x)^{2}}\)
(d) \(\frac{3-x}{(1-x)^{2}}\)

Answer/Explanation

Answer: (d) \(\frac{3-x}{(1-x)^{2}}\)
Explanation:
Given, an = 2n + 3
Since generating function for 1 is \(\frac{1}{1-x}\) and n is \(\frac{x}{(1-x)^{2}}\), the generating function for an is
A(x) = \(\frac{2 x}{(1-x)^{2}}\) + \(\frac{3}{1-x}\)
= \(\frac{2 x+3(1-x)}{(1-x)^{2}}\) = \(\frac{3-x}{(1-x)^{2}}\)
which is option (d).


Question 40.
Let U = {1,2, ….., n} and A = {(x , X), x ∈ X and X ⊆ U}. Consider the following two statements om |A|.

Gate Questions on Combinatorics chapter 3 img 2

Which of the following is correct?
(a) Both I and II
(b) Neither I nor II
(c) Only II
(d) Only I

Answer/Explanation

Answer: (a) Both I and II
Explanation:
A – {(x, X), x ∈ X and X ⊆ U}
The number of k element subsets of a set U with n elements = \(\begin{aligned}
&n \\
&k
\end{aligned}\) = nCk
The number of possible ordered pairs (x, X) where x ∈ X is k. nCk for a given value of k from 1 to n.
So total number of ordered pairs in A

Computer Science Combinatorics Questions and Answers chapter 3 img 19

So II is correct, (Note that k = 0 is excluded since the empty set has no elements and cannot form an ordered pair such as (x, X)).
But since by the combinational identity

Computer Science Combinatorics Questions and Answers chapter 3 img 20

So I am also correct.
So both I and II are correct.


Question 41.
The number of permutations of the characters in LILAC so that no character appears in its original position, if the two L’s are indistinguishable, is_________.

Answer/Explanation

Explanation:
Since both L’s are indistinguishable.
First L’s can be arranged in 3 positions 2, 3, or 5 in 3C2 = 3 ways as follows:
_ L _ L _
or _ L _ _ L
or _ _ _ L L
Now the letters I, A, C can be deranged in 2 x 2! ways. Example in _ L _ L _. C cannot occupy 5th position, so only 2 ways.
The remaining I and A can be arranged in the remaining 2 positions in 2! ways = 2 ways.
So answer is 3 × 2 × 2! = 12.


Question 42.
There are 6 jobs with distinct difficulty levels, and 3 computers with distinct processing speeds. Each job is assigned to a computer such that:

• The fastest computer gets the toughest job and the slowest computer gets the easiest job.
• Every computer gets at least one job.

The number of ways in which this can be done is ______.

Answer/Explanation

Explanation:
Let computers be A, B, and C
The toughest job assigned to the fastest computer (Say, A) is 1 way.
The easiest job assigned to the shortest computer (Say, B) is 1 way.
The remaining 4 jobs are to be assigned to 3 computers so that computer C gets at least one job since A and B have already been assigned a job.
A number of ways 4 jobs assigned to 3 computers = 34.
The number of ways 4 jobs assigned to 3 computers so that computer C does not get any job = 24. Required number of ways = 34 – 24 = 81 – 16 = 65 ways


Set Theory and Algebra Questions and Answers | Computer Science Quiz

Computer Science Set Theory and Algebra MCQ Quiz Questions and Answers PDF Download

Computer Science Set Theory and Algebra Questions and Answers

Question 1.
State whether the following statements are TRUE or FALSE:
The union of two equivalence relations is also an equivalence relation.

Answer/Explanation

Explanation:
A relation is said to be equivalence relation is

  • Reflexive
  • Symmetric, and
  • Transitive

The union of two reflexive relations and two symmetric relations are reflexive and symmetric respectively. However, the union of two transitive relations need not be transitive. Therefore, the union of two equivalence relations need not be an equivalence relation.
Example:
Let R1 and R2 on set A = {1, 2, 3}
R1 = {(1, 1), (2, 2,), (3, 3) (1, 2), (2, 1)} is an equivalence relation
R2 = {(1, 1),(2, 2), (3, 3), (2, 3), (3, 2)} is an equivalence relation
R1 ∪ R2 = {(1, 1), (2, 2), (3, 3,), (1, 2), (2, 1), (2, 3), (3,2)} is not an equivalence relation, because (1, 2) & (2, 3) needs (1, 3) element to be in transitive relation.


Question 2.
(a) How many binary relations are there on a set A with n elements?
(b) How many one—to—one functions are there from a set A with n elements onto itself.

Answer/Explanation

Explanation: (a) Set A contains n elements. Every subset of A × A is a binary relation on set A.
∴ Number of binary relations on a set A with n elements 2
(b) Number of one-to-one functions from a set A with m-elements to a set B with n – elements are nPm So the number of one-to-one functions from a set A with n-elements to itself is nPn = n!.


Question 3.
The complement(s) of the element ‘a’ in the lattice shown in the figure is (are)

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 1

Answer/Explanation

Explanation: The complement of an element x is x’ iff LUB of x and x’ is 1 (greatest element) and GLB of x and x’ is 0 (least element).
∴ The complement of the element ‘a’ in the lattice = {b, c, d, e}.


Question 4.
The transitive closure of the relation {(1, 2) (2, 3) (3, 4) (5, 4)} on the set A = {1, 2, 3, 4, 5} is _______.

Answer/Explanation

Explanation:
The relation has (1, 2) (2, 3) then add (1, 3) to the relation. Since relation have (2, 3) (3,4) then add (2, 4) to relation.
So the resultant relation is = {(1, 2), (2, 3), (3, 4), (5, 4), (1, 3), (2, 4)} Now the resultant relation have (1,2) (2, 4) then add (1, 4) to the relation
∴ {(1,2), (2,3), (3,4), (5,4), (1,3), (2,4), (1,4)}
So the transitive closure of the relation is = {(1,2), (2,3), (3,4), (5,4), (1,3), (2,4), (1,4)}


Question 5.
Let S be an infinite set and S1, S2, … Sn sets such that S1 ∪ S2 ∪ ……. ∪ Sn = S. Then
(a) At least one of the sets Si is a finite set
(b) Not more than one of the sets Si can be finite
(c) At least one of sets Si is infinite
(d) Not more than one of the sets Si is an infinite set.

Answer/Explanation

Answer: (c) At least one of sets S; is infinite
Explanation:
S = S1 ∪ S2 ∪ S3 ∪ … ∪ Sn.
For S to be an infinite set, at least one of sets Si must be infinite, since if all Si were finite, then S will also be finite.


Computer Science Set Theory and Algebra Questions and Answers

Question 6.
Let A be a finite set of size n. The number of elements in the power set of A × A is
(a) 22ⁿ
(b) 2
(c) 2n
(d) 2n

Answer/Explanation

Answer: (b) 2
Explanation:
Number of elements in A × A = n2
∴ A number of elements in the power set of A × A = 2.


Question 7.
Some group (G, o) is known to be abelian. Then, which one of the following is true for G?
(a) g = g-1 for every g∈ G.
(b) g = g2 for every g ∈ G
(c) (goh)2 = g2 o h2 for every g, h ∈ G.
(d) G is of finite order

Answer/Explanation

Answer: (c) (goh)2 = g2 o h2 for every g, h ∈ G.
Explanation:
(goh)2 = (goh)o(goh)
Since group is abelian so it is commutative as well as associative.
= go (hog)h
= go (goh) oh
= (gog) o(hh) = g2 o h2
So (goh)2 = g2 o h2 for every g, h∈ G


Question 8.
Let R be asymmetric and transitive relation on a set A. Then
(a) R is reflexive and hence an equivalence relation
(b) R is reflexive and hence a partial order
(c) R is reflexive and hence not an equivalence relation
(d) None of the above

Answer/Explanation

Answer: (d) None of the above
Explanation:
A relation that is symmetric and transitive, need not be reflexive relation.

  • R = { }; on the set A = {a, b}. The relation R is symmetric and transitive but not reflexive.
  • R = {(a, a), (b, b)}; on the set A = {a, b}. The relation R is symmetric, transitive and also reflexive.

∴ A relation is transitive and symmetric relation but need not be reflexive relation.


Question 9.
The number of elements in the power set P(S) of the set S ={(Φ), 1, (2, 3)} is:
(a) 2
(b) 4
(c) 8
(d) None of these

Answer/Explanation

Answer: (c) 8
Explanation:
If a set has n elements then its powers set has 2n elements.
Given set S = {(Φ), 1, (2, 3)}
Number of elements in S = 3
∴ The number of elements in powerset (S)
= 23 = 8.


Question 10.
Let A be the set of all nonsingular matrices over real numbers and let * be the matrix multiplication operator. Then
(a) A is closed under * but <A, *> is not a semigroup.
(b) <A, *> is a semigroup but not a monoid.
(c) <A, *> is a monoid but not a group.
(d) <A, *> is a group but not an abelian group

Answer/Explanation

Answer: (d) <A, *> is a group but not an abelian group
Explanation:

  • Closure Property: Multiplication of two non-singular matrices is also a non-singular matrix. Matrix multiplication over non-singular matrices follows closure properties.
  • Associative Property: Multi-plication over any set of matrices is associative. (AB)C = A(BC) Where A, B, and C are non-singular matrices
  • Identity Element: Identity matrix I am the identity element for matrix multiplication over matrices and I is non-singular.
  • Inverse Element: For every non-singular matrix its inverse exists. So, for non¬singular matrices, inverse elements exist.
  • Commutative: Matrix multiplication is not commutative
    AB ≠ BA
    Where A, B are non-singular matrices. Matrices multiplication is not commutative. So <A, *> is a group but not an abelian group

Computer Science Set Theory and Algebra Questions and Answers

Question 11.
Let A and B set and let Ac and Bc denote the complements of the sets A and B. The set (A – B) ∪ (B – A) ∪ (A n B) is equal to
(a) A ∪ B
(b) Ac ∪ Bc
(c) A ∩ B
(d) Ac ∩ Bc

Answer/Explanation

Answer: (a) A ∪ B
Explanation:
(A – B) ∪ (B – A) ∪ (A ∩ B)
Representing the above set using the Venn diagram as follows.

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 10

 

∴ By Venn diagram,
(A – B) ∪ (B – A) ∪ (A ∩ B) = A ∪ B


Question 12.
Let X = {2, 3, 6,12, 24}. Let ≤ be the partial order defined by x ≤ y if x divides y. The number of edges in the Hasse diagram of (X, ≤) is
(a) 3
(b) 4
(c) 9
(d) None of these

Answer/Explanation

Answer: (b) 4
Explanation:
x = {2, 3, 6, 12, 24}
The Hasse diagram of (x, ≤) is

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 11

 

Therefore, the number of edges in the Hasse diagram = 4


Question 13.
Suppose X and Y are sets and |x| and |Y| are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. From this one can conclude that
(a) |X| = 1, |Y| =97
(b) IXj =97, |Y| = 1
(c) j X | = 97, | Y | = 97
(d) None of the above

Answer/Explanation

Answer: (a) |X| = 1, |Y| =97
Explanation:
X and Y are set. The cardinalities of X and Y are |X| and |Y| respectively.
The number of functions from X to Y = (|Y|)|X| Given that number of functions from X to Y = 97
∴ 97 = (|Y|)|X|
So above implies that |X| = 1 and |Y| = 97


Question 14.
Which of the following statements is false?
(a) The set of rational numbers is an abelian group under addition
(b) The set of integers in an abelian group under addition
(c) The set of rational numbers form an abelian group under multiplication
(d) The set of real numbers excluding zero in an abelian group under multiplication

Answer/Explanation

Answer: (c) The set of rational numbers form an abelian group under multiplication
Explanation:
0 is also a rational number and for 0, the inverse doesn’t exist under multiplication. Therefore, the set of rational numbers doesn’t form an abelian group under multiplication.


Computer Science Set Theory and Algebra Questions and Answers

Question 15.
Let R denote the set of real numbers. Let f:R x R → R x R be a bijective function defined by f(x, y) = (x + y, x – y). The inverse function of f is given by
(a) f-1(x,y) = \(\left(\frac{1}{x+y}, \frac{1}{x-y}\right)\)
(b) f-1(x,y) = (x – y, x + y)
(c) f-1(x,y) = \(\left(\frac{x+y}{2}, \frac{x-y}{2}\right)\)
(d) f-1(x,y) = (2(x – y), 2(x + y))

Answer/Explanation

Answer: (c) f-1(x,y) = \(\left(\frac{x+y}{2}, \frac{x-y}{2}\right)\)
Explanation:
f(x, y) = ((x + y), (x — y))
So let z1 = x+ y ……(i)
z2=x—y …(ii)
f(x, y) = (z1, z2)
So f-1(z1, z2) = (x, y)
Adding (i) and (ii)
z1 + z2 = 2x
X = \(\frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\)
Subtracting (i) and (ii)
z1 – z2 = 2y
y = \(\mathrm{y}=\frac{\mathrm{z}_{1}-\mathrm{z}_{2}}{2}\)
So the inverse function of F is given by
f-1(z1 , z2) = (x,y) = \(\left(\frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}, \frac{\mathrm{z}_{1}-\mathrm{z}_{2}}{2}\right)\)
Since z1 and z2 are just dummy variables so replacing z1 and z2 by x and y respectively
f-1(x,y) = \(\left(\frac{x+y}{2}, \frac{x-y}{2}\right)\)

Alternate Method:

Given f(x, y) = ((x + y), (x — y))
Take some random ordered pair say (2, 3) f(2, 3) ((2 + 3), (2 — 3)) = (5, —1)
Now the correct inverse must map (5. —1) back to(2,3).
Trying the options one by one we find that only option (e) maps (5, -1) to (2, 3).


Question 16.
Let R be a non-empty relation on a collection of sets defined by ARB if and only if A ∩ B =Φ. Then, (pick the true statement)
(a) R is reflexive and transitive
(b) R is symmetric and not transitive
(c) R is an equivalence relation
(d) R is not reflexive and not symmetric

Answer/Explanation

Answer: (b) R is symmetric and not transitive
Explanation:
(i) Reflexive
A ∩ A = A ≠ Φ
So; (A, A) doesn’t belong to relation R.
∴ Relation R is not reflexive.

(ii) Symmetric
If A ∩ B = Φ then B ∩ A Φ is also true
∴ relation R is a symmetric relation.

(iii) Transitive
If A ∩ B = Φ and B ∩ C = Φ. it need not be true that A ∩ C = Φ.
For example:
A = {1, 2}, B = {3, 4}, C = {1, 5, 6}
A ∩ B = Φ and B ∩ C = Φ but
A ∩ C = {1} ≠ Ø
∴ Relation R is not transitive relation.


Question 17.
Which one of the following is false?
(a) The set of all bijective functions on a finite set forms a group under function composition.
(b) The set {1, 2, …, p – 1} forms a group under multiplication mod p where p is a prime number.
(c) The set of all strings over a finite alphabet Σ forms a group under concatenation.
(d) A subset s ≠ Φ of G is a subgroup of the group <G, *> if and only if for any pair of elements a, b ∈ s, a * b-1∈ s.

Answer/Explanation

Answer: (c) The set of all strings over a finite alphabet Σ forms a group under concatenation.
Explanation:
In option (c). the set of all strings over a finite alphabet Σ doesn’t form a group under concatenation because the inverse of a string doesn’t exist with respect to concatenation.


Computer Science Set Theory and Algebra Questions and Answers

Question 18.
The number of equivalence relations on the set {1, 2, 3, 4} is
(a) 15
(b) 16
(c) 24
(d) 4

Answer/Explanation

Answer: (a) 15
Explanation:
Corresponding to every partition of the set 1, 2, 3, 4, there exists a unique equivalence relation. So we count every type of unordered partition of the set of 4 elements into one block, two-block, three-block, and four block partitions. as shown below:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 12

 

Total 1+7+6+1=15
So the number of equivalence relations on the set {1, 2, 3. 4} is 15.


Question 19.
Suppose A is a finite set with n elements. The number of elements in the Largest equivalence relation of A is
(a) n
(b) n2
(c) 1
(d) n + 1

Answer/Explanation

Answer: (b) n2
Explanation:
A is a finite set with n elements. The largest equivalence relation of A is the cross. product A × A and the number of elements in the cross product A × A is n2.


Question 20.
Let R1 and R2 be two equivalence relations on a set. Consider the following assertions:
(i) R1 ∪ R2 is an equivalence relation
(ii) R1 ∩ R2 is an equivalence relation
Which of the following is correct?
(a) both assertions are true
(b) assertion (i) is true but assertion (ii) is not true
(c) assertion (ii) is true but assertion (i) is not true
(d) neither (i) nor (ii) is true

Answer/Explanation

Answer: (c) assertion (ii) is true but assertion (i) is not true
Explanation:
A relation is said to be an equivalence relation if the relation is
(i) Reflexive
(ii) Symmetric
(iii) Transitive
Reflexive and symmetric properties are both closed under ∪ & ∩.
Transitive property is closed under n but not u. So equivalence relations are closed under ∩ hut not ∪.
Therefore R1 ∩ R2 is an equivalence relation but R1 ∪ R2 is not necessarily an equivalence relation.


Computer Science Set Theory and Algebra Questions and Answers

Question 21.
The number of functions from an m element set to an n element set is
(a) m + n
(b) mn
(c) nm
(d) m * n

Answer/Explanation

Answer: (c) nm
Explanation:
The number of functions from an m element set to an n clement set is nm.


Question 22.
The binary relation R = {(1, 1), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4)} on the set A = {1, 2, 3, 4} is
(a) Reflexive, symmetric and transitive
(b) Neither reflexive, nor irreflexive but transitive
(c) Irreflexive, symmetric and transitive
(d) Irreflexive and antisymmetric

Answer/Explanation

Answer: (b) Neither reflexive nor irreflexive but transitive
Explanation:
The relation R doesn’t contain (4, 4), so R is not reflexive relation. Since relation R contains (1, 1), (2, 2), and (3, 3). Therefore, relation R is also not irreflexive.
That R is transitive, can be checked by systematically checking for all (a. b) and (b, c) in R, whether (a, e) also exists in R. So option (b) is correct.


Question 23.
Suppose A = {a, b, c, d} and Π1 is the following partition of A
Π1 = {{a, b, c}, {d}}
(a) List the ordered pairs of the equivalence relations induced by Π1
(b) Draw the graph of the above equivalence relation

Answer/Explanation

Answer: Given partition of A is
π1 = {(a, b, c), (d)}
(a) The ordered pairs of the equivalence relations induced by π1 is
R = (a, b, c) × (a, b, c) ∪ (d) × (d)
R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c), (d, d)}
(b) The digraph of the above equivalence relation is as follows:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 13


Question 24.
Let (A, *) be a semigroup. Furthermore, for every a and b in A, if a ≠ b, then a * b ≠ b * a.
(a) Show that for every an in A a * a = a
(b) Show that for every a, b in A a * b * a = a
(c) Show that for every a, b, c in A a * b * c = a * c

Answer/Explanation

Answer:
Explanation:
Given a ≠ b ⇒ a * b ≠ b * a The contrapositive version of this is a*b = b*a ⇒ a = b
(a) Now since (A, *) is given to be a semigroup, * is associative
i. e. a * (a * a) = (a * a) * a
Hence a = a * a
(b) (a * b * a) * a = a * b * (a * a) = a * b * a
⇒ a*(a*b*a) = (a*a)*b*a = a * b * a
as a * a = a [proved in part a]
So (a * b * a) * a = a * (a * b * a)
Hence a * b * a = a
(c) (a * b * c) * (a * c) = a * b * (c * a * c) = a * b * c
(a * c) * (a * b * c) = (a * c * a) * b * c = a * b * c
So (a *b * c) * (a * c) = (a * c) * (a * b * c)
Hence a*b*c = a*c


Computer Science Set Theory and Algebra Questions and Answers

Question 25.
The number of binary relations on a set with n elements is
(a) n2
(b) 2n
(c) 2ⁿ²
(d) None of these

Answer/Explanation

Answer: (c) 2ⁿ²
Explanation:
The maximum number of elements in a binary relation on a set A with n elements = Number of elements in A × A = n2
Each element has two choices, either to appear on a binary relation or doesn’t appear on a binary relation.
∴ The number of binary relations = 2.


Question 26.
(a) Mr. X claims the following:
If a relation R is both symmetric and transitive, then R is reflexive. For this, Mr. X offers the following proof. “From xRy, using symmetry we get yRx. Now because R is transitive, xRy and yRx together imply xRx. Therefore, R is reflexive.” Briefly point out the flaw in Mr. X’s proof.
(b) Give an example of a relation R which is symmetric and transitive but not reflexive.

Answer/Explanation

Explanation:
(a) According to Mr. X if xRy is in relation then yRx is also in relation because of symmetricity. Now because R is transitive, xRy and yRx implies xRx.
The flaw in Mr. X claim is that if xRy is present then only it implies yRx and hence xRx. However, if xRy is not present then according to symmetricity yRx is also not present. So xRx need not be present in relation. So it need not be reflexive.
For Example, Empty relation Φ is both symmetric and transitive. However, Φ is not reflexive relation.
(b) Examples of relations that are symmetric and transitive but not reflexive.
(i) Empty relation Φ is symmetric and transitive but not reflexive.
(ii) Relation R = {(a, b), (b, a), (a, a), (b, b)} over A = {a, b, c} is both symmetric and transitive but not reflexive.


Question 27.
A relation R is defined on the set of integers as xRy iff (x + y) is even. Which of the following statements is true?
(a) R is not an equivalence relation
(b) R is an equivalence relation having 1 equivalence class
(c) R is an equivalence relation having 2 equivalence classes
(d) R is an equivalence relation having 3 equivalence classes

Answer/Explanation

Answer: (c) R is an equivalence relation having 2 equivalence classes
Explanation:
A relation R is defined as xRy iff (x + y) is even over a set of integers.
(x + y) is even iff
(i) both x and y are even
(ii) both x and y are odd
Therefore, relation R is equivalence relation because relation is
(i) Reflexive
x + x = 2x = even
So (x, x) belongs to R. So relation is reflexive.
(ii) Symmetric
If x + y = even then y + x is also even So relation is symmetric.
(iii) Transitive
If x + y = even and y + z = even
Then x + y + y + z = even + even
⇒ x + z + 2y = even
⇒ x + z = even – 2y
⇒ x + z = even
∴ Relation R is transitive.
So relation R is an equivalence relation which divides the set of integer into two equivalence classes: One is of all even and other is of odd integer.

Equivalence classes of R are
[0] = { ….. -6, -4, -2, 0, 2, 4, 6, …}
[1] = { ….. -7, -5, -3, -1, 1, 3, 5, 7, …}


Question 28.
Let P(S) denote the power set of a set S. Which of the following is always true?
(a) P(P(S)) = P(S)
(b) P(S) ∩ P(P(S)) = {Ø}
(c) P(S) ∩ S = P(S)
(d) S ∉ P(S)

Answer/Explanation

Answer: (b) P(S) ∩ P(P(S)) = {Ø}
Explanation:
Φ always present in any powerset of a set and Φ is the only common element between P(S) and P(P(S))
∴ P(S) ∩ P(P(S)) = {Φ}


Question 29.
Consider the following relations
R1(a, b) iff (a + b) is even over the set of integers
R2(a, b) iff (a + b) is odd over the set of integers
R3(a, b) iff a.b > 0 over the set of non-zero rational numbers.
R4 (a, b) iff | a – b | ≤ 2 over the set of natural numbers
Which of the following statements is correct?
(a) R1 and R2 are equivalence relations, R3 and R4 are not
(b) R1 and R3 are equivalence relations, R2 and R4 are not
(c) R1 and R4 are equivalence relations, R2 and R3 are not
(d) R1, R2 R3, and R4 are all equivalence relations

Answer/Explanation

Answer: (b) R1 and R3 are equivalence relations, R2 and R4 are not
Explanation:
(I) Relation R1(a, b) iff (a + b) is even over the set of integers.
(i) a + a = 2a which is even So (a, a) belongs to R1
∴ R1 is reflexive relation

(ii) If (a + b) is even, then (b + a) is also even
∴ R1 is symmetric relation

(iii) If (a + b) and (b + c) are even then a + c = (a + b) + (b + c) – 2b = even + even — even = even
∴ R1 is transitive relation.
Since R1 is reflexive, symmetric and transitive so R1 is an equivalence relation.

(II) RR2(a, b) iff (a +b) is odd over set of integers,
(i) a + a = 2a which is not odd So (a, a) doesn’t belong to R2
∴ R2 is not reflexive relation Since R2 is not reflexive, it is not an equivalence relation.

(III) R3(a, b) iff a . b > 0 over set of non-zero relational numbers.
(i) a. a. > 0 for every non-zero rational number.
∴ R3 is reflexive relation.
(ii) If a.b > 0 then b.a > 0
∴ R3 is symmetric relation.
(iii) a.b > 0 and b.c > 0 ⇒ All a,b,c are positive or all a,b,c are negative.
So a.c > 0
∴ R3 is transitive relation.
So R3 is an equivalence relation.

(IV) R4 (a, b) iff |a – b| ≤ 2 over the set of natural number
(i) |a – a| ≤ 2
0 ≤ 2
∴ R4 is reflexive relation.

(ii) If |a — b| ≤ 2 then also |b – a| ≤ 2 ∴ R4 is symmetric relation
(iii) If |a — b| ≤ 2 and |b — c| ≤ 2 then it is not necessary that |a — c| ≤ 2
Ex. |3 – 5| ≤ 2 and |5 — 7| ≤ 2 but |3 —7| ≤ 2
∴ R4 is not transitive.
Since R4 is reflexive and symmetric not transitive, so R4 is not an equivalence relation.


Question 30.
Consider the following statements:
S1: There exist infinite sets A, B, C such that A ∩ (B ∩ C) is finite.
S2: There exist two irrational numbers x and y such that (x + y) is rational.
Which of the following is true about S1 and S2?
(a) Only S1 is correct
(b) Only S2 is correct
(c) Both S1 and S2 are correct
(d) None of S1 and S2 is correct

Answer/Explanation

Answer: (c) Both S1 and S2 are correct
Explanation:
S1: Let A = set of integers
B = set of odd integers
C = set of even integers
A ∩ (B ∩ C) = Φ and Φ is finite set.
Therefore, S1 is true.
S2: Let two irrational number x and y are respectively \((1+\sqrt{2})\) and \((1-\sqrt{2})\).
So x + y= 1 + \(\sqrt{2}\) +1 – \(\sqrt{2}\)
= 2 which is rational number Therefore, S2 is true. Since both S1 and S2 are true, option (c) is true.


Computer Science Set Theory and Algebra Questions and Answers

Question 31.
Let f: A → B be a function and let E and F be subsets of A. Consider the following statements about images.
S1:f(E ∪ F)=  f(E) ∪ f(F)
S2: f(E ∩ F) = f(E) ∩ f(F)
(a) Only S1 is correct
(b) Only S2 is correct
(c) Both S1 and S2 are correct
(d) None of S1 and S2 is correct

Answer/Explanation

Answer: (a) Only S1 is correct
Explanation:
Given a function f: x → y and subsets E and F of A then we have f(E ∪ F) = f(E) ∪ f(F) and f(E ∩ F) ⊆ f(E) ∩ f(F)
Therefore S1 is correct and S2 is false.


Question 32.
The binary relation S = f(empty set) on set A = {1, 2, 3} is
(a) Neither reflexive nor symmetric
(b) Symmetric and reflexive
(c) Transitive and reflexive
(d) Transitive and symmetric

Answer/Explanation

Answer: (d) Transitive and symmetric
Explanation:
The empty relation on any set is always transitive and symmetric but not reflexive.


Question 33.
Consider the set Σ* of all strings over the alphabet Σ = {0,1}. Σ* with the concatenation operator for strings
(a) does not form a group
(b) forms a non-commutative group
(c) does not have a right identity element
(d) forms a group if the empty string is removed from Σ*

Answer/Explanation

Answer: (a) does not form a group
Explanation:
Σ = {0, 1}
Σ* = {0,1}*
= {ε, 0, 1, 01, 10, 11, 000, …}
So (Σ*, •) is an algebraic system, where • (concatenation) is a binary operation.
So (Σ*, •) is a group if and only if the following conditions are satisfied.
1. • (Concatenation) is a closed operation.
2. • is an associative operation.
3. There is an identity.
4. Every element of Σ* has an inverse

Condition 1: * is a closed operation because for any ω1,∈ Σ* and ω2 ∈ Σ*, ω1 . ω2∈ Σ*
Condition 2: For any string x, y, z ∈ Σ*, x . (y . z) = (x . y) . z
So it is associative for example let
x = 01, y = 11, z = 00 then L.H.S. = x . (y.z)
= 01(1100) = 01(1100) = 011100 R.H.S. = (x.y).z
= (0111)00 = (0111)00 = 011100
Condition 3: The Identity is e or empty string because for any string ω ∈ Σ*,
εω = ωε = ω
Now, since ε ∈ Σ*, identity exists.
Condition 4: There is no inverse exist for Σ* because any string ω ∈ Σ*, there is no string ω-1 such that ω . ω-1 = ε = ω-1 ω.
So Σ* with the concatenation operator for strings doesn’t form a group but it does form a monoid.


Question 34.
Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S →{True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y e S satisfying x ≤ y, where ⇒ stands for logical implication. Which of the following statements CANNOT be true?
(a) P(x) = True for all x ∈ S such that x ≠ b
(b) P(x) = False for all x ∈ S such that x ≠ a and x ≠ c
(c) P(x) = False for all x ∈ S such that b ≤ x such that x ≠ c
(d) P(x) = False for all x ∈ S such that a ≤ x and b ≤ x

Answer/Explanation

Answer: (d) P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
Explanation:
If a ≤ x, since p(x) ⇒ p(y) whenever x ≤ y ∴ p(a) ⇒ p(x)
Now since p(a) = True, p(x) = cannot be false. ∴ (d) cannot be true.


Computer Science Set Theory and Algebra Questions and Answers

Question 35.
Consider the set {a, b, c} with binary operators + and x defined as follows :

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 2

For example, a + c = c, c + a = a, c x b = c and b x c = a.
Given the following set of equations:
(a x x) + (a x y) = c
(b x x) + (c x y) = c
The number of solution (s) (i.e., pair (s) (x, y) that satisfies the equations) is
(a) 0
(b) 1
(c) 2
(d) 3

Answer/Explanation

Answer: (c) 2
Explanation:
The possible solution pairs are (a, a), (a, b), (a, e), (b, a), (b, b), (b, c), (c, a,), (c, b) and (e, c). Substitute them one by one in both equations and see which of them satisfies both the equations.
The given equations are:
(a × x)+(a × y) = C ……(i)
(b × x)+(c × y) = C ……(ii)
Substitute first (x. y) = (a, a)
LHS of equation (i) becomes (a × a) + (a × a) = a +a = b
Now RHS of equation (i) = e
Therefore LHS ≠ RHS. This means that (a, a) is not a solution pair. Similarly, try each of the remaining seven possible solution pairs. It will be found that only two pairs (b, c) and (c, b) will satisfy both equations (i) & (ii) simultaneously. Therefore choice (c) is correct.


Question 36.
Consider the binary relation:
S = {(x, y) | y = x + 1 and x, y e {0, 1, 2, ….}}
The reflexive transitive closure of S is
(a) {(x, y) | y > x and x, y ∈ {0, 1, 2, ….}}
(b) {(x, y) | y ≥ x and x,y ∈ {0, 1, 2, …..}}
(c) {(x, y) | y < x and x,y ∈ {0, 1, 2, ….}}
(d) {(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ….}}

Answer/Explanation

Answer: (b) {(x, y) | y ≥ x and x,y ∈ {0, 1, 2, …..}}
Explanation:
S = {(x, y) | y = x + 1, and x, y ∈ (0, 1, 2, …}} = {(0, 1), (1, 2), (2, 3), (3, 4), …}
Now let T1 be the reflexive closure of S.
T1 = {(0, 0), (1, 1), (2, 2), (3, 3)…} ∪ {(0, 1), (1, 2), (2, 3), (3, 4)…}
= {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 4) …}

Let T2 be the transitive closure of S.
(0, 1), (1, 2) ∈ S ⇒ (0, 2) ∈ T2
(0, 2), (2, 3) ∈ S ⇒ (0, 3) ∈ T2
(0, 3), (3, 4) ∈ S ⇒ (0, 4) ∈ T2
and so on…..
Also (1, 2), (2, 3) ∈ S ⇒ (1, 3) ∈ T2
(1, 3), (3, 4) ∈ S ⇒ (1, 4) ∈ T2
(1, 4), (4, 5) ∈ S ⇒ (1, 5) ∈ T2
and so on….
∴ T2 = {(0, 1), (0, 2), (0, 3), …, (1, 2), (1, 3), (1, 4), …}

Now the reflexive, transitive closure of S will be T3 = T1 ∪ T2 = {(0, 0),(0, 1), (0, 2),…, (1,1), (1, 2), (1, 3),…,(2, 2) (2, 3), (2, 4),…}.
Option (b) is correct.


Question 37.
The number of different nxn symmetric matrices with each element being either 0 or 1 is: (Note : power (2, x) is same as 2x)
(a) power(2, n)
(b) power(2, n2)
(c) power(2, (n2 + n)/2)
(d) power (2, (n2 – n)/2)

Answer/Explanation

Answer: (c) power(2, (n2 + n)/2)
Explanation:
In a symmetric matrix, the lower triangle must be the mirror image of the upper triangle using the diagonal as a mirror. Diagonal elements may be anything. Therefore, when we are counting symmetric matrices we count how many ways are there to fill the upper triangle and diagonal elements. Since the first row has n elements, second (n -1) elements, third row (n – 2) elements, and so on up to the last row, one element.
Total number of elements in diagonal + upper triangle
= n + (n – 1) + (n – 2) + … + 1 = n(n + 1)/2
Now, each one of these elements can be either 0 or 1. So total number of ways we can fill these elements is
\(2 \frac{n(n+1)}{2}\) = power (2, (n2 + n)/2)
Since there is no choice for lower triangle elements the answer is power (2, (n2 + n)/2) which is choice (c).


Question 38.
In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses. How many students have not taken any of the three courses?
(a) 15
(b) 20
(c) 25
(d) 30

Answer/Explanation

Answer: (c) 25
Explanation:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 14

According to inclusion – exclusion formula:
| PL ∪ DS ∪ CO | = | PL | + | DS | + | CO | – | PL ∩ DS|- |PL – CO|-|DS ∩ CO| + |PL ∩ DS ∩ CO| PL = students who have taken programming. DS = Students who have taken Data structures. CO = Students who have taken Computer Organization.
So, the number of students who have taken atleast 1 of the 3 courses is given by:
= 125 + 85 + 65- 50- 35- 30 + 15 = 175
Therefore, the number of students who have not taken any of the 3 courses is = Total students – students taking at least 1 course = 200 – 175 = 25


Computer Science Set Theory and Algebra Questions and Answers

Question 39.
Let R1 be a relation from A = {1, 3, 5, 7} to B= {2, 4, 6, 8} and R2 be another relation from B to C = {1, 2, 3, 4} as defined below:
(i) An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
(ii) An element x in B is related to an element y in C (under R2) if x + y is even but not divisible by 3.
Which is the composite relation R1R2 from A to C?
(a) R1R2={(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)}
(b) R1R2={1, 2), (1, 3), (3, 2), (5, 2), (7, 3)}
(c) R1R2={1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
(d) R1R2={(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)}

Answer/Explanation

Answer: (c) R1R2={1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
Explanation:
R1 = {(An element x in A, An element y in B) if x + y divisible by 3}
R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}
R2 = {(An element x in B, An element y in C) if x + y is even but not divisible by 3}
R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}
So, R1 o R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}.


Question 40.
The following is the incomplete operation table of a 4-element group.

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 3

The last row of the table is
(a) c a e b
(b) c b a e
(c) c b e a
(d) c e a b

Answer/Explanation

Answer: (d) c e a b
Explanation:
Step 1:
By looking at the row for e, we see that it is a copy of the column headers. So e must be the identity element. Since right identity and left identity elements must both be the same. The column corresponding to e must be a copy of the row headers. We can now say that the operation table is:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 15

 

Step 2:
From the table above we see that a*c = e
∴ c*a must also be = e (if a is the inverse of c, then c is the inverse of a)
Now the operation table looks like

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 16

 

Step 3:
The blank in the second column must he c (since in a Cayley table, every row and every column is a unique permutation of the row and column headers).
Now the operation table looks like

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 17

 

Step 4:
Now the blanks in the third row can be filled as a, e or e, a. Let us try each one in turn. If we fill a, e in the third row the operation table will look like

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 18

 

Now the blank in the fourth row and third column must be filled by e. However, this is not possible since e is already entered in the fourth row and second column. Therefore filling a, e in third-row blanks is wrong. So let us try filling the third-row blanks with e, a Now the operation table looks like

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 19

 

Now the blanks in the fourth row have to be filled with a, b The final operation table looks like

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 20

 

which is consistent with all the rules of a Cayley Table. The last row of this table is c, e, a, b Therefore the correct answer is (d).


Question 41.
The inclusion of which of the following sets into S = {{1, 2}, {1,2,3}, {1, 3, 5}, {1, 2, 4}, {1, 2, 3, 4, 5}} is necessary and sufficient to make S a complete lattice under the partial order defined by set containment?
(A) {1}
(b) {1}, {2, 3}
(c) {1}, {1, 3}
(d) {1}, (1, 3}, {1, 2, 3, 4}, {1, 2, 3, 5}

Answer/Explanation

Answer: (A) {1}
Explanation:
The hasse diagram of the given post is:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 21

 

In a complete lattice L, every nonempty subset of L has both LUB and GLB. Now it is necessary to add {1} since GLB of {1,2} and {1, 3, 5} is {1}. The hasse diagram now becomes:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 22

 

Now the above hasse diagram represents a complete lattice since every nonempty subset has both LUB and GLB. Therefore adding {1} is not only necessary but it is also sufficient to make the given lattice a complete lattice. Therefore the correct choice is (a).


Question 42.
Let A, B and C be non-empty sets and let X = (A – B) – C and Y = (A – C) – (B – C) Which one of the following is TRUE?
(a) X = Y
(b)X ⊂ Y
(c) Y ⊂ X
(d) None of these

Answer/Explanation

Answer: (a) X = Y
Explanation:
X = (A-B) – C
= (A ∩ B’) – C
= (A ∩ B’) ∩ C’
= AB’C’
Y= (A-C)-(B-C)
= (A ∩ C) – (B ∩ C’)
= (AC’) – (BC’)
= (AC’) ∩ (BC’)’
= (AC’) ∩ (B’ + C)
= (AC’) . (B’ + C)
= AC’B’ + AC’C
= AC’B’ (Since C’C = 0)
= AB’C’ (commutative property)
∴ X =Y


Computer Science Set Theory and Algebra Questions and Answers

Question 43.
The following is the Hasse diagram of the poset [{a, b, c, d,e}, ≤]

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 4

The poset is
(a) not a lattice
(b) a lattice but not a distributive lattice
(c) a distributive lattice but not a Boolean algebra
(d) a Boolean algebra

Answer/Explanation

Answer:(b) a lattice but not a distributive lattice
Explanation:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 23

 

The poset [{a, b, c, d, e}, ≤ ] is a lattice (since every pair of elements has LUB and GLB) but it is not a distributive lattice. Because distributive lattice satisfies the following conditions. For any x, y, z,
x ∧ (y v z) — (x ∧ y) v (x ∧ z)
x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)
Where ∧ and ∨ are meet and join operations but for given poset [{a, b, c, d, e} ≤]
b ∧ (c ∨ d) = b ∧ a = b
(b ∧ c) ∨ (b ∧ d) = e ∨ e = e
So it is not distributive. (Also, element b has 2 complements c and d, which is not possible in a distributive lattice, since, in a distributive lattice, complement if it exists, is always unique).


Question 44.
The set {1, 2, 4, 7, 8,11,13,14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively
(a) 3 and 13
(b) 2 and 11
(c) 4 and 13
(d) 8 and 14

Answer/Explanation

Answer: (c) 4 and 13
Explanation:
The set S = {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15.
The identity element for this group is e = 1 since,
∀x ∈ S, 1 . x mod 15 = x
Now let the inverse of 4 be 4-1.
Now (4 . 4-1) mod 15 = e = 1
Since (4 . 4) mod 15=1
∴ 4-1 = 4 (This inverse is unique)
Similarly let the inverse of 7 be 7-1
(7 . 7-1) mod 15=1
putting each element of set as 7-1 by trial and error we get
(7 . 13) mod 15 = 91 mod 15=1
∴ 7-1 = 13
So, 4-1 and 7-1 are respectively 4 and 13. The correct choice is (c).


Question 45.
Let R and S be any two equivalence relations on a non-empty set A. Which one of the following statements is TRUE?
(a) R ∩ S, R ∪ S are both equivalence relations
(b) R ∪ S is an equivalence relation
(c) R ∩ S is an equivalence relation
(d) Neither R ∪ S nor R ∩ S is an equivalence relation

Answer/Explanation

Answer: (c) R ∩ S is an equivalence relation
Explanation:
R ∩ S is an equivalence relation as can be seen from proof given below.
Let ∀x ∈ A (x, x) ∈ R and (x, x) ∈ S (since R and S are reflexive)
∴ (x, x) ∈ R n S also ∴ R ∩ S is reflexive.
Now, (x, y) ∈ R ∩ S
⇒ (x, y) ∈ R and (x, y) ∈ S
⇒ (y, x) ∈ R and (y, x) ∈ S
(Since R and S are symmetric)
⇒ (y,x) ∈ R ∩ S
∴ (x, y) ∈ R ∩ S
⇒ (y, x) ∈ R ∩ S
R ∩ S is therefore symmetric Now consider
(x, y) and (y, z) ∈ R ∩ S ⇒ (x, y) and (y, z) ∈ R and (x, y) and (y, z) ∈ S ⇒ (x, z) ∈ R and (x, z) ∈ S
(Since R and S are transitive)
⇒ (x, z) ∈ R ∩ S
∴ R ∩ S is transitive also. Since R ∩ S is reflexive, symmetric, and transitive.
∴ R ∩ S is an equivalence relation.
Note: A similar argument cannot be made from R ∪ S.


Question 46.
Let f: B → C and g: A → B be two functions and let h = f o g. Given that h is an onto function. Which one of the following is TRUE?
(a) f and g should both be onto functions
(b) f should be onto but g need not be onto
(c) g should be onto but f not be onto
(d) both f and g need not be onto

Answer/Explanation

Answer: (b) f should be onto but g need not be onto
Explanation:
Consider the arrow diagram shown below:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 24

 

h(a) = f . g(a) = α
h(b) = f . g(b) = β
Here f is onto but g is not onto, yet h is onto. As can be seen from the diagram if f is not onto, h cannot be onto.
∴ f should be onto, but g need not be onto.
∴ The answer is (b).


Computer Science Set Theory and Algebra Questions and Answers

Question 47.
Consider the set H of all 3 × 3 matrices of the type

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 25

 

where a, b, c, d, e, and f are real numbers and abc ≠ 0. Under the matrix multiplication operation, the set H is
(a) a group
(b) a monoid but not a group
(c) a semigroup but not a monoid
(d) neither a group nor a semigroup

Answer/Explanation

Answer: (a) a group
Explanation:
(i) The set H is closed since the multiplication of upper triangular matrices will result only in the upper triangular matrix.
(ii) Matrix multiplication is associative, i.e.
A * (B * C) = (A * B) * C.
(iii) Identity element is

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 26

 

and this belongs to H as I is an upper triangular as well as a lower triangular matrix.

(iv) If A ∈ H, then | A | = abc. Since it is given that abc ≠ 0, this means that |A | ≠ 0 i.e. every matrix belonging to H is non-singular and has a unique inverse.
∴ the set H along with matrix multiplication is a group.


Question 48.
Let f be a function from a set A to a set B, g a function from B to C, and h a function from A to C, such that h(a) = g(f(a)) for all an ∈ A. Which of the following statements is always true for all such functions f and g?
(a) g is onto ⇒ h is onto
(b) h is onto ⇒ f is onto
(c) h is onto ⇒ g is onto
(d) h is onto ⇒ f and g are onto

Answer/Explanation

Answer: (c) h is onto ⇒ g is onto
Explanation:
Given, h = g(f(x)) = g.f
Consider the following arrow diagram:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 27

 

From the above diagram it is clear that g is not onto ⇒ h = g.f is also not onto, since the co-domain of g is the same as the co-domain of g.f. The contrapositive version of the above implication is h is onto ⇒ g is onto which also has to be true since direct = contrapositive. So option (c) is true.


Question 49.
Let A be a set with n elements. Let C be a collection of distinct subsets of A such that for any two subsets S1 and S2 in C, either S1 ⊂ S2 or S2 ⊂ S1 What is the maximum cardinality of C?
(a) n
(b) n+1
(c) 2n-1 +1
(d) n!

Answer/Explanation

Answer: (b) n+1
Explanation:
The way C is defined in the question contains only comparable subsets of A. i.e. the set C is the set of all comparable subsets of set A. Such a set is called a chain. Consider an example with set A = {1, 2} Subsets are Φ, {1}, {2} and {1, 2} Maximum cardinality of collection of distinct subsets is {Φ, {1} and {1, 2}} i.e. 3. So with n elements, the maximum cardinality is n + 1.


Computer Science Set Theory and Algebra Questions and Answers

Question 50.
For the set N of natural numbers and a binary operation f: N × N → N, an element z ∈ N is called an identity for f, if f(a, z) = a = f(z, a), for all an e N. Which of the following binary operations have an identity?
I. f(x, y) = x + y – 3
II. f(x, y) = max(x, y)
III. f(x, y) = xy
(a) I and II only
(b) II and III only
(c) I and III only
(d) None of these

Answer/Explanation

Answer: (d) None of these
Explanation:
I. f(x) = x + y – 3
x + a – 3 = x = a + x – 3
so a = 3
Now 3 is unique, and 3 ∈ N So I has identity.
II: f(x) = max(x, y)
max(x, a) = x = max(a, x)
In N, the only value of a which will satisfy the above equation is a = 1.
Since 3 is unique, and 3 ∈ N So II has an identity.

III. f(x) = xy
xa = x = ax
Now xa = x ⇒ a = 1, but x= ax has no solution for a in the set N.
So III has no identity.
So only I and II have an identity.


Question 51.
Let, X, Y, Z be sets of sizes x, y, and z respectively. Let W = X × Y and E be the set of all subsets of W. The number of functions from Z to E is
(a) z
(b) 2 × 2xy
(c) 2z
(d) 2xyz

Answer/Explanation

Answer: (d) 2xyz
Explanation:
Given | X | = x, | Y | = y and | Z | = z W = X × Y
So | W| = xy
| E | = 2|W| =2xy
So the number of functions for Z to E = | E | |z|
= (2xy)z = 2xyz


Question 52.
The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four possible reasons. Which one of them is false?
(a) It is not closed
(b) 2 does not have an inverse
(c) 3 does not have an inverse
(d) 8 does not have an inverse

Answer/Explanation

Answer: (c) 3 does not have an inverse
Explanation:
Let A = {1, 2, 3, 5, 7, 8, 9}
Construct the table for any x, y ∈ A such that x * y = (x.y) mod 10

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 28

 

We know that 0 ∉ A. So it is not closed. Therefore, (a) is true.
The identity element = 1
∴ (2 . 2-1) mod 10 = 1
From the table, we see that 2-1does not exist. Since, (3 . 7) mod 10 = 1
∴ 7 is the inverse of 3 and 7 ∈ A.
(c) is false.
(d) is true since 8 does not have an inverse.


Question 53.
A relation R is defined on ordered pairs of integers as follows: (x, y)R(u,v) if x < u and y > v. Then R is
(a) Neither a Partial Order nor an Equivalence Relation
(b) A Partial Order but not a Total Order
(c) A Total Order
(d) An Equivalence Relation

Answer/Explanation

Answer: (a) Neither a Partial Order nor an Equivalence Relation
Explanation:
(x, y) R (u, v) iff x < u and y > v
(x, x) (x, x) since x / x and x / x
So R is not reflexive,
∴ R is neither a partial order, nor an equivalence relation.


Question 54.
Let E, F and G be finite sets:
Let X = (E ∩ F) – (F ∩ G) and Y = (E – (E ∩ G)) – (E – F).
Which one of the following is true?
(a) X ⊂ Y
(b) X ⊃ Y
(c) X = Y
(d) X – Y ≠ Ø and Y – X ≠ Ø

Answer/Explanation

Answer: (c) X = Y
Explanation:
Consider the following Venn diagram for X = (E ∩ F) – (F ∩ G)

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 29

Y = (E -(E ∩ G)) – (F – F)

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 30

So, X = Y
or alternatively the solution can be obtained from boolean algebra as follows:
X = (E ∩ F) – (F ∩ G)
= EF – FG
= EF ∩ (FG)’
= EF . (F’ + G’)
= EFF’ + EFG’
= EFG’
Similarly, Y = (E -(E ∩ G)) – (E – F)
= (E – EG) – (E . F’)
= E . (EG)’ – EF’
= E . (E’ + G’) – EF’
= EG’ – EF’
= EG’ . (EF’)’
= EG’ . (E’ + F)
= EE’ G’ + EFG’
= EFG’
Therefore X = Y


Computer Science Set Theory and Algebra Questions and Answers

Question 55.
Let S = {1, 2, 3,…, m}, m > 3. Let X1, X2,…, Xn be subsets of S each of size 3. Define a function f from S to the set of natural numbers as, f(i) is the number of sets Xj that contains the element i. That is f(i) = | {j | i∈ Xj} |.
Then \(\sum_{i=1}^{m}\) (i) is
(a) 3m
(b) 3n
(c) 2m +1
(d) 2n + 1

Answer/Explanation

Answer: (b) 3n
Explanation:
The problem can be solved by considering the cases m = 4 and m = 5 etc.
Let, m = 4
S = {1, 2, 3, 4}
n = number of 3 element subsets
= 4C3 = 4C1 = 4
n = 4
The 4 subsets are {1, 2, 3}, {1, 2, 4}, {1, 3, 4} and {2, 3, 4}
f(1) = number of subsets having 1 as an element = 3
f(2) = number of subsets having 2 as an element = 3
f(3) = 3 and f(4) = 3
∴ \(\sum_{i=1}^{4} f(i)\) = 3 + 3 + 3 + 3=12

Both choice (a) and choice (b) are matching the answer since
3m = 3n = 12
Now let us try m = 5
S = {1, 2, 3, 4, 5}
n = number of 3 element subsets = 5C3 = 10
∴ n = 10
The 10 subsets are {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}
f(1) = f(2) = f(3) = f(4) = f(5) = 6
\(\sum_{i=1}^{5} f(i)\) = 6 + 6 + 6 + 6 + 6 = 30

Clearly 3m = 3 × 5 = 15{is not matching Σf(i)} But
3n = 3 × 10 = 30 {is matching Σf(i) = 30}
∴ 3 n is the only correct answer.
The correct choice is (b).
The problem can also be solved in a more general way as follows:

\(\sum_{i=1}^{m} f(i)\) = f(1) + f(2) + ……+ f(m)
Since, f(1) = f(2) =…….= f(m) = (m-1)C2
= \(\frac{\mathrm{m} \times(\mathrm{m}-1) \times(\mathrm{m}-2)}{2}\)
= \(\frac{3 \times \mathrm{m} \times(\mathrm{m}-1) \times(\mathrm{m}-2)}{1 \times 2 \times 3}\)
= 3 × mC3
Since n = Number of there elements subsets of a set of m elements = mC3
Therefore, \(\sum_{i=1}^{m} f(i)\) = 3 × mC3 = 3n


Question 56.
Let P, Q, and R be set to let △ denote the symmetric difference operator defined as P △ Q = (P ∪ Q) — (P ∩ Q). Using Venn diagrams, determine which of the following is/are TRUE?
I. P △ (Q ∩ R) = (P △ Q) ∩ (P △ R)
II. P ∩ (Q △ R) = (P ∩ Q) △ (P ∩ R)
(a) I only
(b) II only
(c) Neither I nor II
(d) Both I and II

Answer/Explanation

Answer: (b) II only
Explanation:
P △ Q = PQ’ + P’Q
where △ is symmetric difference between P and Q.
I. LHS = P △ Q ∩ R = P△(QR) = P(QR)’ + P'(QR) = P(Q’ + R’) + P’QR
RHS = (P △ Q) ∩ (P △ R) = (PQ’ + P’Q) (PR’ + P’R) = PQ’R’ + P’QR
LHS ≠ RHS So statement I is false.

II. LHS = P ∩ Q △ R = P(QR’ + Q’R)
= PQR’ + PQ’R
RHS = (P ∩ Q) △ (P ∩ R) = PQ △ PR = PQ(PR)’ + (PQ)’ PR = PQ(P’ + R’) + (P’+Q’)PR = PQR’ + PQ’R
LHS = RHS
So statement II is true.


Question 57.
What is the cardinality of the set of integers X defined below? X = {n| 1 ≤ n ≤ 123, n is not divisible by 2, 3 or 5}
(a) 28
(b) 33
(c) 37
(d) 44

Answer/Explanation

Answer: (b) 33
Explanation:
We know that, number of elements divisible by number ‘x’ in set of integer (from 1 to X) = \(\frac{X}{x}\)
So, according to inclusion exclusion formula:
n(2 or 3 or 5) = n(2) + n(3) + n(5) – n(2,3) – n(2,5) – n(3,5) – n (2,3,5))
No’s divisible by 2 in X = \(\left[\frac{123}{2}\right\rfloor\) = 61
No’s divisible by 3 in X = \(\left[\frac{123}{3}\right\rfloor\) = 41
No’s divisible by 5 in X = \(\left[\frac{123}{5}\right\rfloor\) = 24
No’s divisible by 2 and 3 = \(\left.\mid \frac{123}{2 \times 3}\right\rfloor\) = \(\left[\frac{123}{6}\right\rfloor\) = 20
No’s divisible by 3 and 5 = \(\left.\mid \frac{123}{3 \times 5}\right\rfloor\) = \(\left[\frac{123}{15}\right\rfloor\) = 8
No’s divisible by 2 and 5 = \(\)
No’s divisible by 2 and 3 and 5
The problem can be solved by considering the cases m = 4 and m = 5 etc.
= \(\left[\frac{123}{2 \times 3 \times 5}\right\rfloor\) = \(\left[\frac{123}{30}\right\rfloor\) = 4
So, n(2 or 3 or 5) = 61 + 41 +24 — 20 — 8-12 + 4 = 90
n(2 or 3 or 5) = Total number – n(2 or 3 or 5) = 123-90 = 33


Computer Science Set Theory and Algebra Questions and Answers

Question 58.
Let S be a set of n elements. The number of ordered pairs in the largest and the smallest equivalence relations on S is
(a) n and n
(b) n2 and n
(c) n2 and 0
(d) n and 1

Answer/Explanation

Answer: (b) n2 and n
Explanation:
Let S be a set of n elements say {1, 2, 3,…, n}. Now the smallest equivalence relation on S must contain all the reflexive elements {(1, 1), (2, 2), (3, 3),…, (n, n)} and its cardinality is therefore n. The largest equivalence relation on S is S × S, which has a cardinality of n × n = n2.
∴ The largest and smallest equivalence relations on S have cardinalities of n2 and n respectively. The correct choice is (b).


Question 59.
Consider the set S = {a, b, c, d}. Consider the following 4 partitions π12, π3, π4 on
S: π1 = \(\{\overline{a b c d}\}\) , π2 = \(\{\overline{a b}, \overline{c d}\}\) , π3 = \(\{\overline{a b c}, \bar{d}\}\) , π4 = \(\{\bar{a}, \bar{b}, \bar{c}, \bar{d}\}\)
Let ≺ be the partial order on the set of partitions S’ = (π12, π3, π4) defined as follows: πi≺ πj if and only if πi refines πj The poset diagram for (S’, ≺) is

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 5

Answer/Explanation

Answer: (c)
Explanation:
A partition P1 is called a refinement of the partition P2 if every set in P1, is a subset of one of the sets in P2.
π4 is a refinement of π23, and π1
π2 and n3 are refinements of π1
π2 and π3 are not comparable since neither is a refinement of the other.
So the poset diagram for (S’, <) would

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 31

 

Which is choice (c).


Question 60.
A partial order P is defined on the set of natural numbers as follows. Here x/y denotes integer division.
(i) (0, 0) ∈P.
(ii) (a, b) ∈P if and only if a %10 ≤ b %10 and (a/10, b/10) ∈P.
Consider the following ordered pairs:
(i) (101, 22)
(ii) (22, 101)
(iii) (145, 265)
(iv)(0, 153)
Which of these ordered pairs of natural numbers are contained in P?
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (iii) and (iv)

Answer/Explanation

Answer: (d) (iii) and (iv)
Explanation:
In this problem
a % b means the mod function (i.e. residue when a is divided by b).
a/b means integer division (i.e. quotient when a is divided by b)
(i) (101,22):
101% 10 ≤ 22% 10 ⇒ 1 ≤ 2 which is true. (101/10, 22/10) = (10, 2) ∈ P need to check IS (10, 2) ∈ P?
10% 10 ≤ 2% 10 ⇒ 0 ≤ 2 which is True.
Then (10/10, 2/10) = (1, 0) fails since 1 < 0.
∴ (101, 22) ∉ P.

(ii) (22, 101)
22% 10 ≤ 101% 10 ⇒ 2 ≤ 1 is False.
∴ (22, 101)∉ P

(iii) (145, 265)
145% 10 ≤ 265% 10 ⇒ 5 ≤ 5 is true and (145/10, 265/10) = (14, 26) e P has to be checked.
Now consider (14, 26).
14% 10 ≤ 26% 10 ⇒ 4 ≤ 6 is true and (14/10, 26/10) = (1, 2) ∈ P has to be checked. Now consider (1, 2)
1% 10 ≤ 2% 10 ⇒ 1 ≤ 2 is true and (1/10, 2/10) = (0, 0) ∈ P which is given to be true. Therefore (145, 265) ∈ P.

(iv) (0, 153):
0% 10 ≤ 153% 10 ⇒ 0 ≤ 3 is true.
Then (0/10, 153/10) = (0, 15) should be in P. (0, 15):
0% 10 ≤ 15% 10 ⇒ 0 ≤ 5 is true.
Then (0/10, 15/10) = (0, 1) should be in P. (0, 1):
0% 10 ≤ 1% 10 ⇒ 0 ≤ 1 is true.
Then (0/10, 1/10) = (0, 0) should be in P. It is given that (0, 0) ∈ P Therefore (0, 153) ∈ P.
So (iii) and (iv) are contained in P.


Question 61.
If P, Q, R are subsets of the universal set
U, then (P∩Q∩R) ∪ (Pc ∩ Q ∩ R) ∪ Qc ∪ Rc is
(a) Qc ∪ Rc
(b) P ∪ Qc ∪ Rc
(c) Pc ∪ Qc ∪ Rc
(d) U

Answer/Explanation

Answer: (d) U
Explanation:
The given set theory expression can be converted into equivalent boolean algebra expression as follows:
(p∩q∩r) ∪ (pc ∩ q ∩ r) ∪ qc ∪ rc
= p.q.r + p’.q.r + q’ + r’
= qr (p + p’) + q’ + r’
= qr + q’ + r’
= (q + q’) . (r + q’) + r’
= r + q’ + r’
= r + r’ + q’
= 1 + q’
= 1 = U


Computer Science Set Theory and Algebra Questions and Answers

Question 62.
Consider the following Hasse diagrams.

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 6

Which all of the above represent a lattice?
(a) (i) and (iv) only
(b) (ii) and (iii) only
(c) (iii) only
(d) (i), (ii) and (iv) only

Answer/Explanation

Answer: (a) (i) and (iv) only
Explanation:
In lattice, every pair in the Hasse diagram must have Least Upper Bound (LUB) and Greatest Lower Bound (GLB). In figure

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 32

 

For element d, e LUB (d, e) = {b, c} but LUB must be unique i.e. only 1 element present in LUB. Same for GLB (b, c) = {d, e} Hence not lattice.

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 33

 

GLB (b, d) = {c, e} but GLB must be unique i.e. only 1 element present in GLB. Hence not lattice. So, (ii) and (iii) are not lattice


Question 63.
Which one of the following is NOT necessarily a property of a Group?
(a) Commutativity
(b) Associativity
(c) Existence of inverse for every element
(d) Existence of identity

Answer/Explanation

Answer: (a) Commutativity
Explanation:
Group properties are closure, associativity existence of identity, and existence of inverse for every element. Commutativity is not required for a mathematical structure to become a group.


Question 64.
Consider the binary relation: R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?
(a) R is symmetric but NOT antisymmetric
(b) R is NOT symmetric but antisymmetric
(c) R is both symmetric and antisymmetric
(d) R is neither symmetric nor antisymmetric

Answer/Explanation

Answer: (d) R is neither symmetric nor antisymmetric
Explanation:
Given, R {(x, y), (x, z), (z, x), (z, y)) on set {x, y, z}, here (x, y) ∈ R and (y, x) ∉ R.
∴ R is not symmetric
Also(x, z)ER and(z, x) ∈ R.
∴ R is not an autism metric.
R is neither symmetric nor anti-symmetric.


Question 65.
The composition table of a cyclic group is shown below:

 

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 7

Which one of the following choices is correct?
(a) a, b are generators
(b) b, c are generators
(c) c, d are generators
(d) d, a are generators

Answer/Explanation

Answer: (c) c, d are generators
Explanation:
If an element is a generator, all elements must be obtained as powers of that element.
Try a, b, c, d one by one to see which are the generators.
a = a
a2 = a.a = a
a3 = a.a2 = a.a = a and so on.
∴ a is not the generator,
b = b
b2 = b.b = a
b3 = b.b2 = b.a = b
b4 = b.b3 = b.b = a and so on

∴ b is not the generator c = c
c2 c.c = b
c3 = c.c2 = c.b = d
c4 = c.c3 = c.d = a
Since all of a, b, c, d have been generated as powers of c

∴ c is a generator of this group.
Similarly,
d = d
d2 = d.d = b
d3 = d.d2 d.b = c
d4 = d.d3 = d.c a
∴ d is the other generator.


Computer Science Set Theory and Algebra Questions and Answers

Question 66.
What is the possible number of reflexive relations on a set of 5 elements?
(a) 210
(b) 215
(c) 220
(d) 225

Answer/Explanation

Answer: (c) 220
Explanation:
A number of reflexive relations on a set with n elements = 2n² – n.
Here n = 5. So, answer is 25² – 5 = 220.


Question 67.
Consider the set S = {1, ω, ω2}, where ω and ω2 are cube roots of unity. If * denotes the multiplication operation, the structure {S, *} forms
(a) a group
(b) a ring
(c) an integral domain
(d) a field

Answer/Explanation

Answer: (a) a group
Explanation:
The structure ({n, nth roots of unity}, ×) always forms a group. When n = 3 we get the set ({1, w, w2}, ×) which must also form a group.
or
The fact that ({1, w, w2}, × ) forms a group can also be seen by the fact that it satisfies the four group properties.

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 34

 

From the above operation table, we can see that the given operation is closed, on the given set.
2. Associative: “*” is an associative operation.
3. Identity: From the operation table, we can see that the identity element, is “1”.
4. Inverse: From the operation table, we can see that the inverse of 1 is 1, the inverse of w is w2 and the inverse of w2 is w.
So, ({1, w, w2}, *) is a group.
To be a ring, integral domain, or field, we need two binary operations to be specified, whereas here we have only one operation given. So, choice (a) is correct.


Question 68.
How many onto (or surjective) functions are there from an n-element (n ≥ 2) set to a 2-element set?
(a) 2n
(b) 2n – 1
(c) 2n– 2
(d) 2(2n-2)

Answer/Explanation

Answer: (c) 2n– 2
Explanation:
Let n = 2. There are only 2 onto functions as shown below:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 35

 

For, n = 2
Option (a) 2n = 22 = 4
Option (b) 2n – 1 = 22 – 1 = 3
Option (c) 2n — 2 = 22 — 2 = 2
Option (d) 2(2n — 2) = 2(22 – 2) = 4
So only option (c) gives correct answer.
Alternate method:
The number of onto functions from a set A with m elements to set B with n elements where n < m is given by
nm – nC1(n- 1)m+ nC2(n —2)m … + nCn- 1 1m
Here, m = n = 2
So number of onto functions = 2n2C1 1m = 2n – 2
which is choice (c).


Question 69.
A binary operation ⊕ on a set of integers is defined as x ⊕ y = x2 + y2. Which one of the following statements is TRUE about ⊕?
(a) Commutative but not associative
(b) Both commutative and associative
(c) Associative but not commutative
(d) Neither commutative nor associative

Answer/Explanation

Answer: (a) Commutative but not associative
Explanation:
x ⊕ y = x2 + y2
y ⊕ x = y2 + x2
As ‘+’ sign in commutative so x2 + y2 is equal to y2 + x2 so x ⊕ y is commutative.
Now check associativity
x ⊕ (y ⊕ z) = x ⊕ (y2 + z2)
= x2 + (y2 + z2)2
= x2 + y4 + z4 + 2y2z2
(x ⊕ y) ⊕ z = (x2 + y2) ⊕ z
= (x2 + y2)2 + z2
= x4 + y4 + 2x2y2 + z2
x ⊕ (y ⊕ z) t- (x ⊕ y) ⊕ z
So not associative Option (a) is correct.


Computer Science Set Theory and Algebra Questions and Answers

Question 70.
Let S denote the set of all functions f: {0, 1}4 → {0,1}. Denote by N the number of functions from S to the set {0, 1}. The value of log2 log2N is ______.

Answer/Explanation

Explanation:
f : {0, 1}4 → {0,1} ⇒ S is the set of all functions from a 16 element set to a 2 element set.
| S | =216
N = Number of functions from S to 2 element set {0, 1} = 2216 ]
N = 22
∴ log log N = log log 22’6 = log2216 = 16


Question 71.
Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and U of S we say U Consider the following two statements:
S1: There is a subset of S that is larger than every other subset.
S2: There is a subset of S that is smaller than every other subset.
Which one of the following is CORRECT?
(a) Both S1 and S2 are true
(b) S1 is true and S2 is false
(c) S2 is true and S1 is false
(d) Neither S1 nor S2 is true

Answer/Explanation

Answer: (a) Both S1 and S2 are true
Explanation:
Given the following details:
S = {1,2,3,4,. ..2014}
U < V if the minimum element in the symmetric difference of the two sets is in U.
S1: There is a subset of S that is larger than every other subset.
S2: There is a subset of S that is smaller than every other subset.
S1 is true since Φ (which is a subset of S) is larger than every other subset.
i.e. ∀u v < Φ is true since the minimum element in v △ Φ = v, is in v.
(Note: v △ Φ = uΦ + v’Φ = v)
S2 is also true since S (which is a subset of S) is smaller than every other subset, i.e. ∀u S < v is true since the minimum element in S △ v = v’, is in <S.
(Note: S △ v = Sv’ + S’v = lv’ + Ov =v’)
∴ Both S1 and S2 are correct.


Question 72.
Let X and Y be finite sets and f: X → Y is a function. Which one of the following statements is TRUE?
(a) For any subsets A and B of X, | f(A ∪ B)| = |f(A)| + |f(B)|
(b) For any subsets A and B of X, |f(A ∪ B) = |f(A)| + |f(B)|
(c) For any subsets A and B of X, |f(A ∩ B)| = min{|f(A)|,|f(B)|}
(d) For any subsets S and Tof Y, f-1(S ∩ T) = f -1(S) ∩ f-1(T)

Answer/Explanation

Answer: (d) For any subsets S and Tof Y, f-1(S ∩ T) = f -1(S) ∩ f-1(T)
Explanation:
Example:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 36

 

Let A = {1, 2}, B = {3, 4}
(a) |f (A ∪ B)| = |f(A) |+ |f(B)|
⇒ 3 = 2 + 2 is false

(b) f (A ∩ B) = f (A) ∩ f (B)
⇒ Φ = {a} is false

(c) |f(A ∩ B)| =min{|f(A)|, |f(B)|}
⇒ 0 = min (2, 2) = 2 is false

(d) Let S = {a, b}, T- {a, c}
f-1 (S ∩ T)=f-1(S)∩ f-1(T)
⇒ {1, 3} = {1, 2, 3} n {1, 3, 4}
⇒ {1, 3} = {1, 3} is true


Question 73.
Let G be a group with 15 elements. Let L be a subgroup of G. It is known that L ≠ G and that the size of L is at least 4. The size of L is _____.

Answer/Explanation

Explanation:
Order of subgroup divides the order of group (Lagrange’s theorem). So the order of L has to be 1, 3, 5, or 15. As it is given that the subgroup has at least 4 elements and it is not equal to the given group, therefore the order of the subgroup can’t be 1, 3, or 15. Hence it is 5.


Question 74.
Consider the set of all functions f : {0,1, …, 2014} → {0,1, …,2014} such that f(f(i)) = i, for all 0≤i≤ 2014. Consider the following statements:
P: For each such function it must be the case that for every i, f(i) = i.
Q: For each such function it must be the case that for some i, f(i) = i.
R: Each such function must be onto.
Which one of the following is CORRECT?
(a) P, Q, and R are true
(b) Only Q and R are true
(c) Only P and Q are true
(d) Only R is true

Answer/Explanation

Answer: (b) Only Q and R are true
Explanation:
Since it is given that ∀i f(f(i)) = i
It means f . f = I i.e. f = f’-1 That is f is a symmetric function.
Statement P means that every symmetric function is an identity function which is false since there are many symmetric functions other than the identity function.

Example: {(0,1), (1,0), (2, 3), (3,2),…(2012,2013),(2013, 2012), (2014, 2014)} is a symmetric function but not the identity function.

Statement Q means that in every such symmetric function at least one element is mapped to itself and this is true since there is an odd number of elements in the set {0, 1, 2, … 2014}.

Statement R means that every symmetric function is onto which is true since it is impossible to make an into function symmetric. See the below diagram for an into function shown with only 3 elements:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 37

 

In the above function note that (2, 1) exists in the function but not (1,2) and so it is not symmetric.


Question 75.
There are two elements x,y in a group (G, *) such that every element in the group can be written as a product of some number of x’s and y’s in some order. It is known that x*x = y*y = x*y*x*y = y*x*y*x = e where e is the identity element. The maximum number of elements in such a group is______.

Answer/Explanation

Explanation:
(i) e is identity element
(ii) x*x = e, so x = x-1.
(iii) y*y = e, so y = y-1.
(iv) (x*y)*(x*y) = e, so x*y = (x*y)-1 and (y*x)*(y*x) = e, so y*x = (y*x)-1.
Now (x*y)*(y*x) = x*(y*y)*x = x*e*x = x*x = e
So, (x*y)-1. = (y*x)
From (i) and (ii) we get x*y* = y*x
There are only 4 distinct elements possible in this group
1. e
2. x
3. y
4. xy
All other combinations are equal to one of these four as can be seen below:
yx = xy (already proved)
xxx = xe = x
xyy = xe = x
xxy = ey = y
xyx = xxy = y
and so on…….
So the group is G = {e, x, y, x*y
⇒ |G| = 4


Computer Science Set Theory and Algebra Questions and Answers

Question 76.
If g(x) =1 -x and h(x) = \(\frac{x}{x-1}\) , then \(\frac{g(h(x))}{h(g(x))}\) is
(a) \(\frac{h(x)}{g(x)}\)
(b) \(\frac{-1}{x}\)
(c)\(\frac{g(x)}{h(x)}\)
(d) \(\frac{x}{(1-x)^{2}}\)

Answer/Explanation

Answer: (a) \(\frac{h(x)}{g(x)}\)
Explanation:
g(x) = 1 – x, h(x) = \(\frac{x}{x-1}\)
\(\frac{g(h(x))}{h(g(x))}\) = \(\frac{g\left(\frac{x}{x-1}\right)}{h(1-x)}\) = \(\frac{1-\frac{x}{x-1}}{\frac{1-x}{1-x-1}}\) = \(\frac{x-1-x}{\frac{x-1}{\frac{1-x}{-x}}}\) = \(\frac{\frac{-1}{x-1}}{\frac{1-x}{-x}}\) = \(\frac{x}{(x-1)(1-x)}\) = \(\frac{-x}{(1-x)^{2}}\)
\(\frac{h(x)}{g(x)}\) = \(\frac{x}{\frac{x-1}{1-x}}\) = \(\frac{x}{(1-x)(x-1)}\) = \(\frac{-x}{(1-x)^{2}}\)
∴ \(\frac{g(h(x))}{h(g(x))}\) = \(\frac{h(x)}{g(x)}\)


Question 77.
For a set A, the power set of A is denoted by 2A. If A = {5, {6}, {7}}, which of the following options are True?
1. Ø ∈ 2A
2. Ø ⊆ 2A
3. {5,{6}} ∈ 2A
4. {5,{6}} ⊆ 2A
(a) 1 and 3 only
(b) 2 and 3 only
(c) 1, 2 and 3 only
(d) 1, 2 and 4 only

Answer/Explanation

Answer: (c) 1, 2, and 3 only
Explanation:
A = {5, {6}, {7}}
1. Φ ∈ 2A is true. Any power set contains an empty set.
2. Φ ⊆ 2A is true. Empty set is subset of any set.
3. {5, {6}}e 2A is true, power set of A contain {5, {6}} as a 2 element subset of A.
4. {5, {6}} ⊆ 2A is false. Power set of A not contain 5 and {6} as elements.
So {5, {6}} can not be subset of 2A.


Question 78.
Suppose L = {p,q, r, s, t} is a lattice represented by the following Hasse diagram:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 38

For any xy ∈ L, not necessarily distinct, x ∨ y and x ∧ y are join and meet of x, y respectively. Let L<sup>3</sup> = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let pr be the probability that an element (x,y,z) e L<sup>3</sup> chosen equiprobably satisfies x ∨ (y ∧ Z) = (x ∨ y) ∧ (x ∨ z). Then
(a) P<sub>r</sub> = 0
(b) p<sub>r</sub> = 1
(c) 0<p<sub>r</sub>≤\(\frac { 1 }{ 5 }\)
(d) \(\frac { 1 }{ 5 }\) < p<sub>r</sub> < 1

Answer/Explanation

Answer: (d) \(\frac { 1 }{ 5 }\) < p<sub>r</sub> < 1
Explanation:
L has 5 elements
L3 has all ordered triplets of elements of L
⇒ L3 contain 5 × 5 × 5 = 53 = 125 elements.

Gate Questions on Set Theory and Algebra chapter 2 img 38

If q, r, s are chosen, then only it will violate the distributive property. Number of ways to choose q, r, s in any triplet order = 3! = 3 × 2 × 1 = 6.
∴ p(satisfying distributive property)
= 1 – p(violate the distributive property)
= 1 – \(\frac{6}{5^{3}}\) = 0.952 which is between \(\frac{1}{5}\) and 1.


Question 79.
Let R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1. Which one of the following statements about R is True?
(a) R is symmetric and reflexive but not transitive
(b) R is reflexive but not symmetric and not transitive
(c) R is transitive but not reflexive and not symmetric
(d) R is symmetric but not reflexive and not transitive

Answer/Explanation

Answer: (d) R is symmetric but not reflexive and not transitive
Explanation:
aRb iff a and b are distinct a and b have a common divisor other than 1.
(i) R is not reflexive since a and b are distinct i.e, (a, a) ∉ R
(ii) R is symmetric If a and b are distinct and have a common divisor other than 1, then b and a also are distinct and have a common divisor other than 1.
(iii) R is not transitive
If (a, b) ∈ R and (b, c) ∈ R then (a, c) need not be in R.
Example: (2, 6) ∈ R and (6, 2) ∈ R, but (2, 2) ∉ R
∴ R is symmetric but not reflexive and not transitive.


Question 80.
The cardinality of the power set of {0, 1,2…, 10} is _____.

Answer/Explanation

Explanation:
Let X = {0, 1,2,…..10}
|X| = 1
|P(X)| = 211 = 2048


Computer Science Set Theory and Algebra Questions and Answers

Question 81.
The number of onto functions (surjective functions) from set X = {1,2,3,4} to set Y = {a, b, c} is _____.

Answer/Explanation

Explanation:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 39

 

The number of onto functions from A → B where | A | = m and | B | = n is given by the formula
nm – nC1(n – 1)m + nC2(n – 2)m +…
+ (-1)n – 1 nCn – 11m

Here, m = 4 and n = 3
Substituting in above formula we get, 343C1(3 – 1)4 + 3C2 (3 – 2)4 = 81 – 48 + 3 = 36


Question 82.
Let X and Y denote the sets containing 2 and 20 distinct objects respectively and F denote the set of all possible functions defined from X and Y. Let f be randomly chosen from F. The probability of f being one-to-one is _______.

Answer/Explanation

Explanation:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 40

 

Total possible functions = 202 = 400
Number of one-to-one functions = 20P2 = 20 × 19
∴ Required probability = \(\frac{20 \times 19}{20 \times 20}\) = 0.95


Question 83.
Let # be a binary operator defined as X # Y = X’ + Y’ where X and Y are Boolean variables. Consider the following two statements:
S1:(P # Q) # R = P # (Q # R)
S2:Q # R = R # Q
Which of the following is/are true for the Boolean variables P, Q, and R?
(a) Only S1 is True
(b) Only S2 is True
(c) Both S1 and S2 are True
(d) Neither S1 nor S2 is True

Answer/Explanation

Answer: (b) Only S2 is True
Explanation:
X # Y = X’ + Y’
This is the NAND operation. NAND is known to be commutative but not associative. So, only S2 is true.


Question 84.
Suppose U is the power set of the set S = {1,2,3,4,5,6}. For any Te U, let | T| denote the number of elements in T and T’ denote the complement of T. For any T, R ∈ U, let T \ R be the set of all elements in T which are not in R. Which one of the following is true?
(a) ∀X ∈ U(|X| = |X’|)
(b) ∃X ∈ U ∃Y ∈U(|X|= 5, |Y|=5 and X ∩ Y = 0)
(c) ∀X ∈ U ∀Y ∈U(|X |=2,| Y|=3 and Y\F = 0)
(d) ∀X ∈U ∀ Y ∈U(X\Y =Y’\X’)

Answer/Explanation

Answer: (d) ∀X ∈U ∀ Y ∈U(X\Y =Y’\X’)
Explanation:
S = {1, 2, 3, 4, 5, 6}
U is the power set of S ⇒ U = P(S)
U ={{ } {1}, {2}, …{1, 2}, {1, 3}, …{1, 2, 3}, … {1, 2, 3, 4}, …{1, 2, 3, 4, 5} …{1, 2, 3, 4, 5, 6}}
| U |= 26 = 64 elements.
T ∈ U ⇒ T’ ∈ U
R ∈U, T\R=T – R

(a)∀X ∈ U(| X | = | X’ |) means that every subset of S has same size as its complement. Clearly this is False.
(For example, {1} ∈ U, and complement of {1} = {2, 3, 4, 5, 6})
(b)∃X ∈ U∃Y ∈ U(|X|=5,|Y|=5 and X ∩ Y = Φ) means that there are two 5 element subsets of S which have nothing in common. This is clearly False.
Since any two 5 element subsets will have atleast 4 elements in common.
(c) ∀X ∈ U ∀ Y ∈ U(|X| = 2,|y|=3 and X\Y = Φ) means that every 2 element subset X and every 3 element subset Y will have X – Y = Φ i.e., X ⊆ Y.
This is clearly False as can be seen from the example X= {1,2}, Y = {3,4,5} where X’^Y.
(d) ∀X ∈ U ∀T ∈ U(X \Y = Y’\X’) means that for any two subsets X and Y, X\ Y = Y \ X’ i.e. X – Y = Y’ – X’.
This is clearly True since by boolean algebra LHS = X – Y = XY’.
RHS =Y’ -X’ = Y’Xand therefore LHS = RHS.


Computer Science Set Theory and Algebra Questions and Answers

Question 85.
Let R be a relation on the set of ordered pairs of positive integers such that ((p, q), (r, s)) ∈ R if and only if p – s = q — r. Which one of the following is true about R?
(a) Both reflexive and symmetric
(b) Reflexive but not symmetric
(c) Not reflexive but symmetric
(d) Neither reflexive nor symmetric

Answer/Explanation

Answer: (c) Not reflexive but symmetric
Explanation:
R = {(p, q), (r, s) I p – s = q – r}
(i) Check reflexive property
∀(p, q) ∈ Z+ × Z+ ((p, q), (p, q)) ∈ R
is true iff p – q = q – p which is false. So relation R is not reflexive.
(ii) Check symmetry property
If ((p, q), (r, .s’))∈ R then ((r, s), (p, q)) ∈ R
((p, q), (r, s)) ∈ R ⇒ p – s = q – r
((r, s), (p, q)) ∈ R ⇒ r – q = s – p
If p – s = q – r is true, then r — q = s – p is also true by rearranging the equation.
∴ R is symmetric.


Question 86.
A function f : N++ → N+, defined on the set of positive integers N+, satisfies the following properties:
f(n) – f(nl2) if n is even
f(n) = f(n + 5) if n is odd
Let R = {i | ∃ j : f (j) = i} be the set of distinct values that/takes. The maximum possible size of R is _______.

Answer/Explanation

Explanation:
f(n) = f\(\left(\frac{n}{2}\right)\) if n is even
f(n) = f{n + 5) if n is odd
f : N+ → N+
Now f(2) = f\(\left(\frac{2}{2}\right)\) = f(1)
f(3) = f(3 + 5) = f(8) = f\(\left(\frac{8}{2}\right)\) = f(4) = f\(\left(\frac{4}{2}\right)\) = f(2) = (1)
So, f(1) = f(2) = f(3) = f(4) = f(8)
Now let us find f(5) = f(5 +5) = f(10) = f\(\left(\frac{10}{2}\right)\)
= f(5) so f(5) = f(10)
Now let us find f(9)
f(9) = f(9 +5) = f(14) = f\(\left(\frac{14}{2}\right)\) = f(7)
= f(7 + 5) = f(12) = f\(\left(\frac{12}{2}\right)\) = f(6)
= f\(\left(\frac{6}{2}\right)\) = f(3)
So f(9) = f(7) = f(6) = f(3) = f(1) = f(2) = f(4) = f(8)
For n > 10, the function will be equal to one of f(1), f(2)….f(10)
So the maximum no. of distinct values / takes is only 2.
First is f(1) = f(2) = f(3) = f(4) = f(8) = f(9) = f(7) = f(6)
Second is f(5) = f(10)
All other n values will give only one of these two values.


Question 87.
A binary relation R on N × N is defined as follows: (a, b) R (c, d) if a ≤ c or b ≤ d. Consider the following propositions:
P: R is reflexive
Q: R is transitive
Which one of the following statements is TRUE?
(a) Both P and Q are true
(b) P is true and Q are false
(c) P is false and Q are true
(d) Both P and Q are false

Answer/Explanation

Answer: (b) P is true and Q are false
Explanation:
(a, b) R (c, d) if a ≤ c or b ≤ d
P : R is reflexive
Q : R is transitive
Since, (a, b) R (a, b) ⇒ a ≤ a or b ≤ b ⇒ t or t
Which is always True, R is reflexive. Now let us check transitive property
Let(a,b) R (c, d) ⇒ a ≤ c or b ≤ d
and(c, d) R {e, f) ⇒ c ≤ e or d ≤ f
Now let us take a situation
a ≤ c (True) or b ≤ d ( false)
and c ≤ e (False) or d ≤ f (True)
Now we can get neither a ≤ e nor b ≤ f So, (a, b) R (c, d) and (c, d) R (e, f) ≠ (a, b) R (e, f). So clearly R is not transitive.
So Pis true and Q is false. Choice (b) is correct.


Question 88.
Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:
I. Each compound in U \ S reacts with an odd number of compounds.
II. At least one compound in U \S reacts with an odd number of compounds.
Ell. Each compound in U\S reacts with an even number of compounds
Which one of the above statements is ALWAYS TRUE?
(a) Only I
(b) Only II
(c) Only III
(d) None of these

Answer/Explanation

Answer: (b) Only II
Explanation:
This is actually a graph theory Question. Let the components be represented by vertices and if component x reacts with component y then let there be an undirected edge between x and y.Now, 9 of the 23 vertices will have 3 degrees each.
Note: U \ S means U — S, which is the remaining 23 – 9 = 14 vertices.
By Handshaking theorem,
9 × 3 + degree of 14 vertices = 2e = Even. i.e., 27 + degree of 14 vertices = 2e = Even. So, the total degree of 14 vertices must be odd.
Statement I means that the degree of each of the 14 vertices is odd, which means that the total degree of 14 vertices will be even (since even number of odd numbers is always even). But we need the total degree of 14 vertices to be odd so, statement I is false.
Statement III means that everyone of the 14 vertices has even degree, which guarantees that the total degree of 14 vertices will be even. So, statement III is false. Since statement II is the negation of statement III, it must be true.


Question 89.
Consider the set X = {a, b, c, d, ej under the partial ordering:
R = {(a, a), (a, b), (a, c), (a, d), (a, e), (b, b), (b, c), (b, e), (c, c), (c, e), (d, d), (d, e), (e, e)}.
The Hasse diagram of the partial order (X, R) is shown below:

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 9

The minimum number of ordered pairs that need to be added to R to make (X, R) a lattice is ______.

Answer/Explanation

Explanation:
Since the given hasse diagram is already a lattice, three is no need to add any ordered pair. So the answer is 0.


Computer Science Set Theory and Algebra Questions and Answers

Question 90.
The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is ______.

Answer/Explanation

Explanation:
Let A → -5- by 3
B → ÷ by 5
C →÷ by 7
n(A ∪ B ∪ C) = n(A)+n(B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)
= \(\left[\frac{500}{3}\right\rfloor\) + \(\left[\frac{500}{5}\right\rfloor\) + \(\left[\frac{500}{7}\right\rfloor\) – \(\left[\frac{500}{15}\right\rfloor\) – \(\left[\frac{500}{21}\right\rfloor\) – \(\left[\frac{500}{35}\right\rfloor\) + \(\left[\frac{500}{105}\right\rfloor\) = 166 + 100 + 71 – 33 – 23 – 14 + 4 = 271


Question 91.
Let G be a finite group on 84 elements. The size of the largest possible proper subgroup of G is _______.

Answer/Explanation

Explanation:
Given | G | =84
By Lagrang’s theorem any subgroup size is a divisior of 84.
But a proper subgroup cannot have same size as group.
So largest divisor of 84, other than 84 is 42. So, largest proper subgroup can have in size of 42.


Question 92.
Let N be the set of natural numbers. Consider the following sets:
P. Set of Rational numbers (positive and negative).
Q. Set of functions from {0, 1} to N.
R. Set of functions from N to {0, 1}.
S. Set of finite subsets of N.
Which of the sets above is countable?
(a) Q and S only
(b) P and S only
(c) P and R only
(d) P, Q, and S only

Answer/Explanation

Answer: (d) P, Q, and S only

Explanation:
P: Set of rational number → countable
Q: Set of functions from {0, 1} to N → N

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 41

 

0 can be assigned in N ways
1 can be assigned in N ways
There are N × N functions, the cross product of countable set in countable.
R: Set of functions from N to {0,1}

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 42

 

Each of these boxes can be assigned to 0 or 1 so each such function is a binary number with an infinite number of bits.
Example: 0000 is the binary number corresponding to 0 is assigned to all boxes and so on. Since each such binary number represents a subset of N (the set of natural numbers) by characteristic function method, therefore, the set of such function is the same as the power set of N which is uncountable due to Cantor’s theorem, which says that power set of a countably infinite set is always uncountably infinite.
S: Set of finite subsets of N → countably infinite since we are counting only finite subsets. So P, Q, and S are countable.


Question 93.
Let G be an arbitrary group. Consider the following relations on G:
R1: ∀a, b ∈ G, a R1 b if and only if ∃g ∈ G such that a = g-1 bg
R2: ∀a, b ∈ G, a R2 b if and only if a = b-1
Which of the above is/are equivalence relation/ relations?
(a) R1 and R2
(b) R1 only
(c) R2 only
(d) Neither R1 nor R2

Answer/Explanation

Answer: (b) R1 only
Explanation:
R1: ∀a, b ∈ G,a R1 b if and only if ∃g∈ G such that a = g-1bg
Reflexive: a = g-1ag can be satisfied by putting g = e, identity “e” always exists in a group.
So reflexive
Symmetric: aRb ⇒ a = g-1ag for some g
⇒ b = gag-1 = (g-1)-1 ag-1 g-1 always exists for every g ∈ G.
So symmetric
Transitive: aRb and bRc ⇒ a = g1-1bg1 and b = g2-1 cg2 for some g1g2 ∈ G.
Now a=g1-1g2-1cg2g1(g2g1)-1 cg2g1
g1 ∈ G and g2 ∈ G ⇒ g2g1 ∈ G since group is closed so aRb and aRb ⇒ aRc hence transitive
Clearly R1 is equivalence relation.
R2 is not equivalenced it need not even be reflexive, since aR2 a => a = a-1 ∀a which not be true in a group.
R1 is equivalence relation is the correct answer.


Question 94.
Let R be the set of all binary relations on the set {1, 2, 3}. Suppose a relation is chosen from R at random. The probability that the chosen relation is reflexive (round off to 3 decimal places) is _____.

Answer/Explanation

Explanation:
A = {1, 2, 3}
n = |A| = 3
Number of relations on A = 2 = 2 = 29
Number of reflexive relations on A
= 2n² – n  = 23² – 3 = 26


Question 95.
Let G be a group of 35 elements. Then the largest possible size of a subgroup of G other than G itself is _______.

Answer/Explanation

Explanation:
Size of group = O(G) = 35
Lagrange’s theorem says if H is a subgroup of G
then O(H) | 0(G).
So 0(H) can be 1, 5, 7 or 35.
So largest subgroup size other than G itself is 7.


Computer Science Set Theory and Algebra Questions and Answers

Question 96.
A relation R is said to be circular if aRb and bRc together imply cRa.
Which of the following options is/are correct?
(a) If a relation S is reflexive and circular, then S is an equivalence relation.
(b) If a relation S is circular and symmetric, then S is an equivalence relation.
(c) If a relation S is transitive and circular, then S is an equivalence relation.
(d) If a relation S is reflexive and symmetric, then S is an equivalence relation.

Answer/Explanation

Answer: (a) If a relation S is reflexive and circular, then S is an equivalence relation.
Explanation:
Let S be reflexive and circular,
Let us checking symmetry:
Symmetry:
Let xSy
Now since S is reflexive ySy true.
So xSy and ySy is true
Now by circular property we get, ySx
So xSy => ySx
So S is symmetric.
Transitive:
Let xSy and ySz
Now by circular property we get zSx and by symmetry property proved above, we get
zSx => xSz
So xSy and ySz => xSz
So S is transitive.
So S is reflexive, symmetric and transitive and hence an equivalence relation.
So option (a) is true.
Option (b): Let S be circular and symmetric. Let S be defined on set {1, 2, 3}
Now empty relation is circular and symmetric but not reflexive. So S need not be an equivalence relation.
So option (b) is false.
Option (c): Let S be transitive and circular. Let S be defined on the set {1, 2, 3}
Now empty relation again satisfies transitive and circular but is not reflexive. So S need not be an equivalence relation.
So option (c) is false.
Option (d): Reflexive and symmetric need not be transitive for example on {1, 2, 3}.
S = {(L 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
is reflexive and symmetric. But it is not transitive because (1, 2) and (2, 3) belong to S but (1, 3) does not.
So option (d) is false.


Question 97.
Let G be a group of order 6, and Hbe a subgroup of G such that 1 < | H | <6. Which one of the following options is correct?
(a) Both G and H are always cyclic.
(b) G is always cyclic, but H may not be cyclic.
(c) G may not be cyclic, but H is always cyclic.
(d) Both G and H may not be cyclic.

Answer/Explanation

Answer: (c) G may not be cyclic, but H is always cyclic.
Explanation:
|G| = 6
H is subgroup, so by Lagrange’s theorem
| H | = 1, 2, 3 or 6 (Divisor’s of 6) Now it is given that 1 < | H | <6 or | H | = 2 or 3
Since 2 and 3 are both prime and since every group of prime order is cyclic, H is surely cyclic. But order of | G | =6 which is not prime. So G may or may not be cyclic. So G may not be cyclic, but H is always cyclic. Option (c) is correct.


Question 98.
Consider the following sets, where n ≥ 2:
S1: Set of all n × n matrices with entries from the set {a, b, c}
S2: Set of all functions from the set {0,1, 2, n2 – 1} to the set {0, 1, 2}
Which of the following choice(s) is/are correct?
(a) There exists a surjection from S1 to S2.
(b) There does not exist a bijection from S1 to S2.
(c) There does not exist an injection from S1 to s2.
(d) There exists a bijection from S1 to S2.

Answer/Explanation

Answer:
(a) There exists a surjection from S1 to S2.
(d) There exists a bijection from S1 to S2.
Explanation:
| S1 | = 3n
Since each of the n2 entries in n × n matrix can be fills in 3 ways.
|S2| = 3n2
Since | {0, 1, 2} | =3 and | {0, 1, 2, n2 — 1} | = n2)
Now the theorem says A bijection fA → B exists iff |A| = |B|.
Here |S1| = |S2|
So, there has to be a bijection from S1 to S2. So, option (d) is correct. If bijection exists surely surjection also exists. So, option (a) is also correct. Option (b) and (c) are incorrect.


Question 99.
Let S be a set consisting of 10 elements. The number of tuples of the form (A, B) such that A and B are subsets of S and A ⊆ B is _______.

Answer/Explanation

Explanation:
The Venn diagram for this is

Computer Science Set Theory and Algebra Questions and Answers chapter 2 img 43

 

Now every element x in S has only 3 options. It can be x ∈ A or x ∈ B – A or x ∈ S — B. So the number of ways to choose A and B such that A ⊆ B ⊆ S is 310.


Computer Science Set Theory and Algebra Questions and Answers

Question 100.
For two n-dimensional real vectors P and Q, the operation s(P, Q) is defined as follows:
s(P, Q) = \(\sum_{i=1}^{n}\)(P[i].Q[i])

Let L be a set of 10-dimensional non-zero real vectors such that for every pair of distinct vectors P, Q ∈ L, s(P, Q) = 0. What is the maximum cardinality possible for the set L?
(a) 10
(b) 9
(c) 100
(d) 11

Answer/Explanation

Answer: (a) 10
Explanation:
L is the set of 10-dimensional orthogonal vectors.
So cardinality of L ≤ 10.
i.e., Maximum cardinality of L = 10.
So, option (a) is correct.