JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Probability

Let X have a binomial distribution

B (n, p)

such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If

P (X > n - 3) = \frac{k}{2 ^{n}}

, then k is equal to

Answer: (B)

Solution

Let mean

= n p = α

and variance

= n pq = β

. Given

α + β = 24

and

α β = 128

. Solving:

α = 16

β = 8

. Thus

n p = 16

and

n pq = 8

, giving

q = 1/2

, so

p = 1/2

and

n = 32

. Now

P (X > n - 3) = P (X \geq 30) = \frac{1}{2 ^{32}} ((30 32) + (31 32) + (32 32)) = \frac{1}{2 ^{32}} (\frac{32 \times 31}{2} + 32 + 1) = \frac{529}{2 ^{32}}

. Therefore

k = 529

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option