Question
The mean and the variance of a binomial distribution are $$4$$ and $$2$$ respectively. Then the probability of $$2$$ successes is :
A.
$$\frac{{28}}{{256}}$$
B.
$$\frac{{219}}{{256}}$$
C.
$$\frac{{128}}{{256}}$$
D.
$$\frac{{37}}{{256}}$$
Answer :
$$\frac{{28}}{{256}}$$
Solution :
$$\eqalign{
& {\text{Mean}} = np = 4{\text{ and}} \cr
& {\text{Variance}} = npq = 2 \cr
& \therefore \,p = q = \frac{1}{2}{\text{ and }}n = 8 \cr
& \therefore \,p\left( {2{\text{ success}}} \right) = {}^8{C_2}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^2} \cr
& = \frac{{28}}{{{2^8}}} \cr
& = \frac{{28}}{{256}} \cr} $$