Q46 / 90 · JEE Main 2022 (27 Jul Shift 2) · Mathematics

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

A
$528$
Correct answer
B
$529$
C
$629$
D
$630$

Solution

Let $M =$ Mean and $V =$ Variance $(M > V)$
So, $M + V = 24, M V = 128$
$\Rightarrow M = 16$ and $V = 8$
Now, mean $n p = 16$
and variance $n p q = 8 \Rightarrow q = \frac{1}{2}$
$∴ p = \frac{1}{2}, n = 32$
Now $P (X > n - 3) = \frac{1}{2^{n}} ({}^{n}C_{n - 2} + {}^{n}C_{n - 1} + {}^{n}C_{n})$
$∴ \frac{k}{2^{n}} = \frac{1}{2^{n}} ({}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32})$
$\Rightarrow k = \frac{32 \times 31}{2} + 32 + 1$
$= 496 + 33 = 529$

Hence, the correct answer is option: 'B'.

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Solution

Relevant Questions

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