Question

The mean and variance of Binomial Distribution are $4$ and $2$ respectively, then the probability of success is

• A. $\frac{128}{256}$
• B. $\frac{219}{256}$
• C. $\frac{37}{256}$
• D. $\frac{28}{256}$

Answer 1

Answer: (D)

$\frac{28}{256}$

Given $np=4$ and $npq=2,$$q=\frac{npq}{np}=\frac{2}{4}=\frac{1}{2}$ so $p=1-\frac{1}{2}=\frac{1}{2}$

Now $npq=2$

$\therefore n=8$

$\therefore$ BD is given by

$P(X=r){=}^8{C}_r{p}^r{q}^{n-r}$ $\therefore P(X=r=2){=}^8{C}_2{\left(\frac{1}{2}\right)}^8$ $=\frac{28}{256}$