Question

The mean and variance of a binomial distribution are 4 and 2 respectively. Then the probability of exactly two succsses is
(A) \(\frac{7}{64}\)
(B) \(\frac{21}{128}\)
(C) \(\frac{7}{32}\)
(D) \(\frac{9}{32}\)

Answer 1

Ans: (A)
Hint \(: n p=4, n p q=2\)
\(\therefore q=\frac{2}{4}=\frac{1}{2} \Rightarrow p=\frac{1}{2} \quad \therefore n=8\)
\(p(x=2)={ }^{8} C_{2}\left(\frac{1}{2}\right)^{8}=\frac{28}{256}=\frac{7}{64}\)