Question

The mean and the variance in a binomial distribution are found to be $2$ and $1$ respectively. The probability $P(X=0)$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{1}{16}$

Answer 1

The correct option is D $\frac{1}{16}$

Mean of a binomial distribution $=np=2$

Variance of a binomial distribution $npq=1$

$⟹q=\frac{1}{2}$

But $p+q=1$

$⟹p=\frac{1}{2}$

$⟹n=4$

Then, $P(X=0)={}_0^{4}C{\left(\frac{1}{2}\right)}^4=\frac{1}{16}$