JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

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 $P(X>n-3)=\dfrac k{2^n}$ , then k is equal to :

528

529

629

630

Answer and worked solution

Official answer: B

Mean and variance. $m+v=24$ and $mv=128$ , so $m=16$ , $v=8$ : $q=\frac12$ , $n=32$ .

Tail. $P(X\ge30)=\frac{\binom{32}2+32+1}{2^{32}}=\frac{529}{2^{32}}$ .

(B)

Topics:Binomial distribution

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 16