Question

If the mean and variance of a Binomial variate $X$ are $8$ and $4$ respectively, then $P(X<3)$ equals

• A. $\frac{265}{2^{15}}$
• B. $\frac{137}{2^{14}}$
• C. $\frac{137}{2^{16}}$
• D. $\frac{265}{2^{16}}$

Answer 1

The correct option is C

$\frac{137}{2^{16}}$

Explanation of the correct option:

Find the required probability

Given: mean $np=8$, variance $npq=4$

Variance $np(1-p)=4$ $\left[\because q=1-p\right]$

$\Rightarrow$ $8(1-p)=4$

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

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

Since, $np=8$

Thus, $n=16$

We know that $P(X<3)$ can be written as $P(X=0)+P(X=1)+P(X=2)$.

$\begin{array}{rcl}& =& P(X=0)+P(X=1)+P(X=2)\\ & =& {}^{16}C_0{p}^0{q}^{16-0}+{}^{16}C_1{p}^1{q}^{16-1}+{}^{16}C_2{p}^2{q}^{16-2}\\ & =& {}^{16}C_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{16}+{}^{16}C_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{15}+{}^{16}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{14}\left[\because q=1-p\right]\\ & =& \left[{}^{16}C_0+{}^{16}C_1+{}^{16}C_2\right]{\left(\frac{1}{2}\right)}^{16}\\ & =& \left[\frac{16!}{16!\times 1}+\frac{16!}{15!\times 1!}+\frac{16!}{14!\times 2!}\right]\times \frac{1}{2^{16}}\\ & =& \frac{1+16+120}{2^{16}}\\ & =& \frac{137}{2^{16}}\end{array}$

Therefore, $P(X<3)=\frac{137}{2^{16}}$

Hence option $\left(C\right)$ $\frac{137}{2^{16}}$ is the correct answer.