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

$\frac{28}{256}$ $\frac{219}{256}$ $\frac{128}{256}$ $\frac{37}{256}$

$\frac{28}{256}$

$\left.\ \begin{matrix}np = 4\\npq = 2\end{matrix}\right\}⇒ q= \frac{1}{2}, p = \frac{1}{2}, n = 8 $ $p(X=2) = {^8C}_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 = 28. \frac{1}{2^8} = \frac{28}{256}.$