JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Probability

The random variable

X

follows binomial distribution

B (n, p)

for which the difference of the mean and the variance is 1. If

2 P (X = 2) = 3 P (X = 1)

, then

n^{2} P (X > 1)

is equal to

Answer: (C)

Solution

For a binomial distribution

B (n, p)

: mean

= n p

and variance

= n pq

where

q = 1 - p

.

Given:

n p - n pq = 1 \Rightarrow n p (1 - q) = 1 \Rightarrow n p^{2} = 1

Also given:

2 P (X = 2) = 3 P (X = 1)

2 \cdot^{n} C_{2} p^{2} q^{n - 2} = 3 \cdot^{n} C_{1} p q^{n - 1}

2 \cdot \frac{n ( n - 1 )}{2} p^{2} q^{n - 2} = 3 n p q^{n - 1}

n (n - 1) p = 3 n q \Rightarrow (n - 1) p = 3 q = 3 (1 - p)

n p - p = 3 - 3 p \Rightarrow n p = 3 - 2 p

From

n p^{2} = 1

p = \frac{1}{n p}

. Substituting:

n p = 3 - \frac{2}{n p}

Let

n p = k

k = 3 - \frac{2}{k} \Rightarrow k^{2} - 3 k + 2 = 0 \Rightarrow k = 1

k = 2

n p = 1

p^{2} = 1 \Rightarrow p = 1

(impossible). So

n p = 2

and

p = \frac{1}{2}

, giving

n = 4

P (X > 1) = 1 - P (X = 0) - P (X = 1) = 1 - (\frac{1}{2})^{4} - 4 (\frac{1}{2})^{4} = 1 - \frac{5}{16} = \frac{11}{16}

n^{2} P (X > 1) = 16 \cdot \frac{11}{16} = 11

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option