Question

If the mean and variance of a Binomial variable $X$ are $2$ and $1$, respectively, the $P(X\ge 1)$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{4}{5}$
• C. $\frac{7}{8}$
• D. $\frac{15}{16}$

Answer 1

The correct option is D

$\frac{15}{16}$

Explanation of the correct option:

Find the required probability

Given: mean $np=2$, variance $npq=1$

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

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

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

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

Since, $np=2$

Thus, $n=4$

We know that $P(X\ge 1)$ can be written as $1-P(X<1)$.

$\begin{array}{rcl}& =& 1-P(x<1)\\ & =& 1-P(x=0)\\ & =& 1-{}^{4}C_0{p}^0{\left(q\right)}^{4-0}\left[q=1-p\right]\\ & =& 1-\frac{4!}{4!\times 1}\times 1\times {\left(1-\frac{1}{2}\right)}^4\\ & =& 1-{\left(\frac{1}{2}\right)}^4\\ & =& 1-\frac{1}{16}\\ & =& \frac{15}{16}\end{array}$

Therefore, $P(X\ge 1)=\frac{15}{16}$

Hence, option $\left(D\right)$ is the correct answer.