Probability Questions and Answers | Computer Science Quiz

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)

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

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

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 – \(\frac { 25 }{ 36 }\) = \(\frac { 11 }{ 36 }\)

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 = \(\frac { 4 }{ 52 }\) × \(\frac { 3 }{ 51 }\).

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.

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 = ( \(\frac { 3 }{ 8 }\))

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.

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.

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

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 = \(\frac{7}{7^{7}}\) = \(\frac{7}{7^{6}}\)

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
= \(\frac{{ }^{4} C_{1}}{2^{4}}\) + \(\frac{{ }^{4} C_{2}}{2^{4}}\) + \(\frac{{ }^{4} C_{3}}{2^{4}}\)= \(\frac { 7 }{ 8 }\)

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

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).

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
= \(\frac{{ }^{4} \mathrm{C}_{2}}{2^{4}}\) = \(\frac { 6 }{ 16 }\) = \(\frac { 3 }{ 8 }\)

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

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 = \(\frac { 60 }{ 230 }\) = \(\frac { 6 }{ 23 }\)

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)=e^-λ₂x, x > 0
Given Z = min (X, Y)
P(Z ≥ x) = P(X ≥ x, Y ≥ x)
= P(X ≥ x) P(Y ≥ x)
= e^-λ₁x . e^-λ₂x = 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

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

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).

Answer:(a) ⁿC_d/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).

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

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

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

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] = \(\frac { 1 }{ 6 }\)

Answer:(a) 3
Explanation:
E(X) = S(X_i × P_i)
Where X_i = number of tosses until we get successive HEAD or TAIL.
P_i = Probability that we get in X_i 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 n^th toss probability = P(X = n) = 2/2ⁿ So the probability distribution table for X (the number of Tosses) is given below:

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

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:

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

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:

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

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

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 3^rd 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 11^th 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.

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

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)

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).

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:

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

Answer:(a) 3
Explanation:
Given µ_X = 1, σ_x² = 4 => σ_x = 2 and µ_Y = -1.σ_Y is unknow
Given, p(X ≤ -1) = p(Y ≥ 2)
Converting into standard normal variates

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 ⇒σ_y = 3

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

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

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

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 2^a . 5^b 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 2^a. 5^b 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)
= \(\frac { 4 × 4 }{ 100 × 100 }\) = \(\frac { 1 }{ 625 }\)

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 = \(\frac { 1/4 }{ 3/4 }\) = \(\frac { 1 }{ 3 }\)

Answer:
(c) R ≥ 0
Explanation:
V(x) = E(x²) – [E(x)]² = R
where V(x) is the variance of x,
Since variance is σ²_x and hence never negative, R ≥ 0.

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

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 I^st card and II^nd card we have to take ordered pairs]
p(I^st card = II^nd card + 1)
= P{(2, 1), (3, 2), (4, 3), (5, 4)} = \(\frac { 4 }{ 5 × 4 }\) = \(\frac { 1 }{ 5 }\)

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

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

So the correct option is (c).

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:

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

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.

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

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

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

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

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

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 = \(\frac { 1 }{ 2 }\)
y_max = \(\frac { 1 }{ 2 }\) – (\(\frac { 1 }{ 2 }\))² = 0.25

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

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

Explanation:

Explanation:

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)

Explanation:

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

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.

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(\(\bar{P}\) | \(\bar{Q}\) = \(\frac { 4 }{ 5 }\)

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

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

Explanation:
P(one of them wins in 3^rd trial)
= P(I^st trial is Tie) × P(II^nd trial is Tie) × P(one of them wins 3^rd 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)

Explanation:
The condition probability table given is

P(H_G) = 0.2, P(M_G) = 0.5, P(L_G) = 0.3 Drawing the tree diagram for HD we get,

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

Explanation:

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

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

Explanation:

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

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.

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

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.

If the received signal is H, the probability that the transmitted signal was H (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

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.

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.

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