Question

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
• B. 529
• C. 629
• D. 630

Answer 1

Answer: (B)

529

Answer 2

The correct answer is (B) : 529
Mean = np = 16
Variance = npq = 8
$⇒ q = p = \frac{1}{2}$ and $n = 32$
$P(x>n-3) = p(x = n-2) + p(x = n-1) + p(x = n)$
$= (^{32}C_2 + ^{32}C_1 + ^{32}C_0). \frac{1}{2^n}$
$= \frac{529}{2^n}$