Question

If the mean and variance of a random variable $X$ having a Binomial distribution are $4$ and $2$, respectively, then find the value of $P(X=1)$

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

Answer 1

The correct option is D

$\frac{1}{32}$

Explanation of the correct option:

Find the required probability

Given: mean $np=4$, variance $npq=2$

Variance $np(1-p)=2$ $\left[\because q=1-p\right]$

$\Rightarrow$ $4(1-p)=2$

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

$\Rightarrow$ $p=\frac{1}{2}$

Since, $np=4$

Thus, $n=8$

We know that $P(X=1)={}^{n}C_1{p}^1{q}^{n-1}$ [Binomial theorem]

$\begin{array}{rcl}& \Rightarrow & P(x=1)={}^{8}C_1{p}^1{q}^{4-1}\\ & \Rightarrow & P(x=1)={}^{8}C_1{\left(\frac{1}{2}\right)}^1{\left(1-\frac{1}{2}\right)}^{8-1}\left[\because q=1-p\right]\\ & \Rightarrow & P(x=1)=\frac{8!}{7!\times 1}\times \frac{1}{2}\times {\left(\frac{1}{2}\right)}^7\\ & \Rightarrow & P(x=1)=\frac{8}{2^8}\\ & \Rightarrow & P(x=1)=\frac{8}{256}\\ & \Rightarrow & P(x=1)=\frac{1}{32}\end{array}$

Therefore, $P(X=1)=\frac{1}{32}$

Hence, option $\left(D\right)$ $\frac{1}{32}$ is the correct answer.