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

Given that, mean$=4\Rightarrow np=4$ and variance = 2 $\Rightarrow$               $npq=2$ $\Rightarrow$               $4q=2$ $\Rightarrow$               $q=\frac{1}{2}$ $\therefore$  $p=1-q=1-\frac{1}{2}=\frac{1}{2}$ Also, $n=8$ Probability of 2 successes$=P(X=2){{=}^{8}}{{C}_{2}}{{p}^{2}}{{q}^{6}}$ $=\frac{8!}{2!6!}\times {{\left( \frac{1}{2} \right)}^{2}}\times {{\left( \frac{1}{2} \right)}^{6}}=28\times \frac{1}{{{2}^{8}}}=\frac{28}{256}$