In a ward-robe, Nitish has 3 trousers. One of them is black, second is blue and third brown. In this ward-robe, he has 4 shirts also. One of them is black and the other 3 are white. He opens his ward-robe in the dark and picks out one shirt-trouser pair without examining the colour. What is the likelihood that neither the shirts nor the trousers are black?

Here, total balls are 14.
? Required probability = (³c₁ + ⁷c₁) / ¹⁴c₁
= (3 + 7)/14
= 10/14
= 5/7

Let the number of green marble = x
Then, Probability of getting a green marble
= ^xc₁ / ²⁴c₁ = 2/3
? x/24 = 2/3
? x = 16

Let P(B) = x
Given, P(A?B) = 0.8 and P(A) = 0.3
? P(A) + P(B) - P(A?B) = 0.8
? P(A) + P(B) - P(A) P(B) = 0.8 {?A and B are independent}
? 0.3 + x - 0.3x = 0.8
? 0.7x = 0.5
? x = 5/7

Total events = n (s) = 2⁵ = 32
n(E) of getting heads = 1
p(E) = 1/32
? n(E) = 1 - p(E) = 1 - 1/32 = 31/32

Total ways = ¹³c₂
Favourable number ways of selecting men only = ⁸c₂
? Probability of selecting no woman
= ⁸c₂ / ¹³c₂
= 14/39
? Probability of selecting at least one woman
= 1 - (14 / 39)
= 25 / 39

Since, there are only two types of socks in the bag. so, if murari pick up 3 socks then certainly two of them are of same type. Thus this is a certain event.
Hence required probability = 1

First we find the probability that the plane is not hit
= (1 - 0.4) ( 1 - 0.3) (1 - 0.2) (1 - 0.1)
= 0.6 x 0.7 x 0.8 x 0.9 = 0.3024

? probability that the plane is hit = 1 - 0.3024
= 0.6976

Total number of possible arrangement for 4 boy and 3 girls in a queue = 7!
when they occupy alternate position then the arrangement would be like
BGBGBGB. Thus the total number of possible arrangement = 4! x 3!
? Required probability = 4! x 3! / 7!
= (4 x 3 x 2 x 3 x 2) / (7 x 6 x 5 x 4 x 3 x 2) = 1/35

Required probability = favorable number of cases/ total number of
=⁸P₅ / 8 ⁵ = 105/512

Total number of caps = 12
n(s) = ¹²c₁ = 12
Out of ( 2 blue + 1 yellow) caps, number of ways to pick one cap n(E) = ³c₁ = 3
? Required probability = p(E) = n (E) / n(s) = 3/12 = 1/4