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

$\frac{1}{32}$

Let n and p be the parameters of the binomial distribution.

We have,

Mean = 4 and Variance = 2

$⇒np = 4  \, and \,  npq = 2 $

$⇒p =\frac{1}{2}= q \, and \, n = 8 $

∴ Required probability $=P(X=1)={^8C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7=\frac{1}{32}$