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 $24\text{and}128$ respectively. If $P(X>n-3)=\frac{k}{2^n}$ , then $k$ is equal to :

Answer 1

$\text{Mean}np=16$
$\text{Variance}=npq=8$
$\Rightarrow q=p=\frac{1}{2}\text{and}n=32$
$P(x>n-3)=p(x=n-2)+p(x=n-1)+p(x=n)$
$=({}^{32}{C}_2+{}^{32}{C}_1+{}^{32}{C}_0)\cdot \frac{1}{2^n}$
$=\frac{529}{2^n}$