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 :
mean = $$np = 4$$ and variance = $$npq = 2$$
$$\eqalign{
& \therefore \,\,p = q = \frac{1}{2}\,\,{\text{and }}n = 8 \cr
& \therefore \,\,P\left( {2\,\,{\text{success}}} \right) = {\,^8}{C_2}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^2} \cr
& = \frac{{28}}{{{2^8}}} \cr
& = \frac{{28}}{{256}} \cr} $$