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 [2023]

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Solution

Q.
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 [2023]

1 15

2 11

3 16

4 12

Ans.
(2)

$n p - n p q = 1$

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

$\Rightarrow n p^{2} = 1$    $[∵ p = 1 - q]$ ...(i)

$Now, 2 P (X = 2) = 3 P (X = 1)$ ...(ii)

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

$\Rightarrow n p - p = 3 q$    $[∵ q = 1 - p]$

$\Rightarrow n p + 2 p = 3$ $[from (i) and (ii)]$

$\Rightarrow p = \frac{1}{2}$

$From (i), n p^{2} = 1$

$\Rightarrow n \times {(\frac{1}{2})}^{2} = 1 \Rightarrow n = 4$

$P (X > 1) = 1 - P (X = 0) + P (X = 1)$

$= 1 - [4 C_{0} {(\frac{1}{2})}^{4} + 4 C_{1} (\frac{1}{2}) {(\frac{1}{2})}^{3}] = \frac{11}{16}$

$∴$    $n^{2} p (x > 1) = 4^{2} \times \frac{11}{16} = 11$

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