Q.The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is (A) $\dfrac{128}{256}$ (B) $\dfrac{219}{256}$ (C) $\dfrac{37}{256}$ (D) $\dfrac{28}{256}$

For $X\sim B(n,p)$, three summary numbers describe the whole distribution without listing every term of the p.m.f.: the mean $\mu=E(X)=np$ (the average number of successes over many repetitions of the $n$-trial experiment), the variance $\text{Var}(X)=npq$ (how spread out the successes are around that mean), and the standard deviation $SD(X)=\sigma_X=\sqrt{npq}$. These are stated as formulae without proof. They are used two ways: forward, when $n$ and $p$ are known, to compute $E(X)$ and $\text{Var}(X)$ directly; and backward, when two of $E(X)$, $\text{Var}(X)$, $n$, $p$ are given and the other two must be recovered - typically by dividing $\text{Var}(X)=npq$ by $E(X)=np$ to isolate $q=\text{Var}(X)/E(X)$ first (since the $n$ cancels), then $p=1-q$, and finally $n=E(X)/p$.