Question 1

# A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is 6/25 12/5 6 4

Solution

B.

12/5

We can apply binomial probability distribution Variance = npq
5

Question 2

## A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends,3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is 484 485 468 469

Solution

B.

485

Total number of ways
4C0 · 3C3 · 3C3 · 4C0 + 4C1 · 3C2 · 3C2 · 4C1 + 4C2 · 3C1 · 3C1 · 4C2 + 4C3 · 3C0 · 3C0 · 4C3= 485

Question 3

## A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is 1/729 8/9 8/729 8/243

Solution

D.

8/243

Probability of getting score 9 in a single throw = 4/36 = 1/9
Probability of getting score 9 exactly twice =

Question 4

Solution

C.

Question 5

## A random variable X has the probability distribution:X: 1 2 3 4 5 6 7 8 P(X): 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 For the events E = {X is a prime number} and F = {X < 4}, the probability P (E ∪ F) is 0.87 0.77 0.35 0.50

Solution

B.

0.77

E = {x is a prime number} = {2, 3, 5, 7}
P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62
F = {X < 4}= {1, 2, 3}
P(F) = P(X = 1) + P(X = 2) + P(X = 3)
P(F) = 0.15 + 0.23 + 0.12 = 0.5
E ∩ F = {X is prime number as well as < 4 }
= {2, 3}
P (E ∩ F) = P(X = 2) + P(X = 3)
= 0.23 + 0.12 = 0.35
∴ Required probability
P (E∪ F) = P(E) + P(F) - P(E ∩ F)
P (E∪ F) = 0.62 + 0.5 - 0.35
P (E ∪ F) = 0.77