Question

The mean and the variance of a random variable $X$ having a binomial distribution are $4$ and $2$ respectively, then $P(X=1)$ is

• A. $1/32$
• B. $1/16$
• C. $1/8$
• D. $1/4$

Answer 1

The correct option is A $1/32$

Mean = $np$
$=4$
Variance $=np(1-p)$
$=2$
Hence $4=2(1-p)$
$p=\frac{1}{2}$
Therefore $n=4\left(2\right)$
$=8$
Hence

$P\left(1\right)$$={}^8{C}_1\frac{1}{2^8}$
$=\frac{8}{256}$

$=\frac{1}{32}$