The mean and the variance of a binomial distribution are 4 and 2  respectively. Then, the probability of 2 successes is

$\frac{28}{256}$

Let n and p bet the parameters for the distribution.

We have,

Mean = 4 and Variance = 2

$⇒ np = 4 $ and $ npq = 2 ⇒p = 1 = \frac{1}{2}$ and $ n = 8 $

Let X denote the number of successes. Then,

$P(X=r)={^8C}_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{8-1}={^8C}_r  \left(\frac{1}{2}\right)^8$

∴ Required probability $= P(X=2)= {^8C}_r  \left(\frac{1}{2}\right)^8= \frac{28}{256}$