Question

A box contains $6$ pens, $2$ of which are defective. Two pens are taken randomly from the box. If r.v. $X$: Number of defective pens obtained, then standard deviation of $X=$

• A. $\pm \frac{4}{3\sqrt{5}}$
• B. $\frac{8}{3}$
• C. $\frac{16}{45}$
• D. $\frac{4}{3\sqrt{5}}$

Answer 1

The correct option is D $\frac{4}{3\sqrt{5}}$

$X:$ no. of defective pens
Two pens are taken from box
$\therefore X$ can take values $0,1,2$
$P(X=0)=\frac{{}^4{C}_2}{{}^6{C}_2}=\frac{4\times 3}{6\times 5}=\frac{2}{5}=\frac{6}{15}$
$P(X=1)=\frac{{}^2{C}_1\times {}^4{C}_1}{{}^6{C}_2}=\frac{2\times 4\times 1\times 1}{6\times 5}=\frac{8}{15}$
$P(X=2)=\frac{{}^2{C}_2}{{}^6{C}_2}=\frac{1\times 2\times 1}{6\times 5}=\frac{1}{15}$

$X$	$P$	$X_i{P}_i$	$X_i^2{P}_i$
$0$	$\frac{6}{15}$	$0$	$0$
$1$	$\frac{8}{15}$	$\frac{8}{15}$	$\frac{8}{15}$
$2$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{4}{15}$

$\sum \left(X_i{P}_i\right)=\frac{10}{15}$
$=\frac{2}{3}$
$\sum \left({X_i}^2{P}_i\right)=\frac{12}{15}$
$=\frac{4}{5}$
Standard deviation $=\sqrt{\sum \left(X_i^2{P}_i\right)-[\sum (X_i{P}_i){]}^2}$
Standard deviation $=\sqrt{\left(\frac{4}{5}\right)-{\left(\frac{2}{3}\right)}^2}$
$=\sqrt{\frac{4}{5}-\frac{4}{9}}$
$=\sqrt{\frac{36-20}{45}}$
$=\frac{4}{3\sqrt{5}}$