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Probability Distribution

Let X have a ...

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 :

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Solution

$Mean n p = 16$
$Variance = n p q = 8$
$\Rightarrow q = p = \frac{1}{2} and n = 32$
$P (x > n - 3) = p (x = n - 2) + p (x = n - 1) + p (x = n)$
$= (^{32} C_{2} +^{32} C_{1} +^{32} C_{0}) \cdot \frac{1}{2^{n}}$
$= \frac{529}{2^{n}}$

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