Question

The mean and the variance of a random variable X having Binomial Distribution are 4 and 2 respectively then

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{32}$

Answer 1

Answer: (D)

$\frac{1}{32}$

Given mean$=4$

Variance$=2$

$\therefore np=4$

and $npq=2$

$\therefore 4q=2,q=\frac{2}{4}=\frac{1}{2}$

$p=1-q=\frac{1}{2}$

$\therefore n\left(1/2\right)=4;n=8$

Thus

$p(X=1){=}^8{C}_1\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^7$

$=8{\left(\frac{1}{2}\right)}^8=\frac{8}{256}=\frac{1}{32}$