Question

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. $\frac{28}{256}$

Answer 1

The correct option is D $\frac{28}{256}$

Given that mean $=4$
$np=4$
and variance $=2$
$npq=2$
$\Rightarrow 4q=2\Rightarrow q=\frac{1}{2}$
$\therefore p=1-q=1-\frac{1}{2}=\frac{1}{2}$ and $n=8$
Probability of 2 successes
$P(X=2)={}^8{C}_2{p}^2{q}^6$
$=\frac{8!}{2!\times 6!}\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^6$
$=28\times \frac{1}{2^8}=\frac{28}{256}$