JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Probability

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If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is:

(A) $\frac{33}{2 ^{32}}$
(B) $\frac{33}{2 ^{29}}$
(C) $\frac{33}{2 ^{28}}$ ✓ Correct
(D) $\frac{33}{2 ^{27}}$

Answer: (C)

Solution

For a binomial distribution with parameters

n

and

p

, mean

= n p

and variance

= n pq

where

q = 1 - p

. Given:

n p + n pq = 24

and

(n p) (n pq) = 128

. Let

n p = x

and

n pq = y

. Then

x + y = 24

and

x y = 128

. Solving:

x

and

y

are roots of

t^{2} - 24 t + 128 = 0

, giving

t = 8

t = 16

. So

n p = 16

and

n pq = 8

, which gives

q = 1/2

, hence

p = 1/2

and

n = 32

. The probability of one or two successes is

P (X = 1) + P (X = 2) = (1 32) (1/2)^{32} + (2 32) (1/2)^{32} = \frac{32 + 496}{2 ^{32}} = \frac{528}{2 ^{32}} = \frac{33}{2 ^{28}}

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Probability

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