Question

An urn contains $4$ white and $3$ red balls.

Three balls are drawn with replacement from this urn.

Then, the standard deviation of the number of red balls drawn is

• A. $\frac{6}{7}$
• B. $\frac{36}{49}$
• C. $\frac{5}{7}$
• D. $\frac{25}{49}$

Answer 1

The correct option is A

$\frac{6}{7}$

Explanation for correct option:

Step1. Finding the probability of red ball

Given, An urn contains $4$ white and $3$ red balls.

Three balls are drawn with replacement from this urn.

Let $X$ represent the number of red balls.

$P$ be the probability of getting red balls $=\frac{3}{7}$

Hence, the probability of not getting red balls,$q=\frac{4}{7}$

⇒ $\begin{array}{rcl}P(X& =& 0)={}^{3}C_0\times {\left(\frac{3}{7}\right)}^0\times {\left(\frac{4}{7}\right)}^3\end{array}$

$\begin{array}{rcl}& =& \frac{64}{343}\end{array}$

⇒ $\begin{array}{rcl}P(X& =& 1)={}^{3}C_1\times {\left(\frac{3}{7}\right)}^1\times {\left(\frac{4}{7}\right)}^2\end{array}$

$\begin{array}{rcl}& =& \frac{144}{343}\end{array}$

⇒ $\begin{array}{rcl}P(X& =& 2)={}^{3}C_2\times {\left(\frac{3}{7}\right)}^2\times {\left(\frac{4}{7}\right)}^1\end{array}$

$\begin{array}{rcl}& =& \frac{108}{343}\end{array}$

⇒ $\begin{array}{rcl}P(X& =& 3)={}^{3}C_3\times {\left(\frac{3}{7}\right)}^3\times {\left(\frac{4}{7}\right)}^0\end{array}$

$\begin{array}{rcl}& =& \frac{27}{343}\end{array}$

Step2. Finding variance

We know that, variance$\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{3}}{\mathit{p}}_{\mathbf{i}}{{\mathbf{x}}_{\mathbf{i}}}^{\mathbf{2}}\mathbf{-}\underset{i=0}{\overset{3}{{[\sum p_i{x}_i]}^2}}$

$\mathrm{X}=$red ball	$\mathrm{P}\left(\mathrm{X}\right)$	$\mathrm{XP}\left(\mathrm{X}\right)$	${\mathrm{X}}^2\mathrm{P}\left(\mathrm{X}\right)$
$0$	$\frac{64}{343}$	$0$	$0$
$1$	$\frac{144}{343}$	$\frac{144}{343}$	$\frac{144}{343}$
$2$	$\frac{108}{343}$	$\frac{216}{343}$	$\frac{432}{343}$
$3$	$\frac{27}{343}$	$\frac{81}{343}$	$\frac{243}{343}$

$\begin{array}{rcl}\sum _{\mathrm{i}=0}^3{p}_{\mathrm{i}}{{\mathrm{x}}_{\mathrm{i}}}^2& =& 0+\frac{144}{343}+\frac{432}{343}+\frac{243}{343}\\ & =& \frac{819}{343}\end{array}$

$\begin{array}{rcl}\underset{\mathrm{i}=0}{\overset{3}{[\sum {\mathrm{p}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}]}}& =& [0+\frac{144}{343}+\frac{216}{343}+\frac{81}{343}]\\ & =& \left[\frac{441}{343}\right]\end{array}$

Put all the value, we get

Variance $=\frac{819}{343}-{\left(\frac{441}{343}\right)}^2$

$=\frac{36}{49}$

Standard deviation $\mathbf{=}\sqrt{\mathbf{v}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}$

Therefore standard deviation $=\frac{6}{7}$

Hence, correct option is $\mathbf{\left(}\mathbf{A}\mathbf{\right)}$