Question

If the mean and the variance of a binomial variate X are 2 and 1 respectively, then probability that X takes a value greater than or equal to one is

• A. $\frac{1}{16}$
• B. $\frac{9}{16}$
• C. $\frac{3}{4}$
• D. $\frac{15}{16}$

Answer 1

The correct option is D $\frac{15}{16}$

For binomial distribution,
Mean $=np=2$ ....(1)
Variance $=npq=1$
$\Rightarrow 2q=1$
$\Rightarrow q=\frac{1}{2}$
$\Rightarrow p=\frac{1}{2}$ ($\because p+q=1$)
$\Rightarrow n=4$
Now, $P(X\ge 1)=1-P(X=0)$
$=1{-}^4{C}_0\left(\frac{1}{2}{)}^0\right(\frac{1}{2}{)}^4$
$=1-\frac{1}{16}$
$=\frac{15}{16}$