Question

There are $\mathrm{n}$ urns each containing $(\mathrm{n}+1)$ balls such that the ${\mathrm{i}}^{\mathrm{th}}$ urn contains $‘\mathrm{i}’$ white balls and $(\mathrm{n}+1-\mathrm{i})$red balls. Let ${\mathrm{u}}_{\mathrm{i}}$ be the event of selecting ${\mathrm{i}}^{\mathrm{th}}$ urn, $\mathrm{i}=1,2,3,\dots ,\mathrm{n}$ and $\mathrm{W}$ denotes the event of getting a white balls. If $\mathrm{n}$ is even and $\mathrm{E}$ denotes the event of choosing even numbered urn $\left[\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)=\frac{1}{\mathrm{n}}\right]$, then the value of $\mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}}\right)$ is

• A. $\frac{\mathrm{n}+2}{2\mathrm{n}+1}$
• B. $\frac{\mathrm{n}+2}{2(\mathrm{n}+1)}$
• C. $\frac{\mathrm{n}}{\mathrm{n}+1}$
• D. $\frac{1}{\mathrm{n}+1}$

Answer 1

The correct option is B

$\frac{\mathrm{n}+2}{2(\mathrm{n}+1)}$

Evaluate the required value based on given information

Given,

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{E}\right)& =& \mathrm{P}\left({\mathrm{u}}_2\right)+\mathrm{P}\left({\mathrm{u}}_4\right)+..............\mathrm{P}\left({\mathrm{u}}_{\mathrm{n}}\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{E}\right)=\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}}+..............\frac{1}{\mathrm{n}}\end{array}$

Since, $\frac{1}{\mathrm{n}}$ is added $\frac{\mathrm{n}}{2}$ times,

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{E}\right)=\frac{1}{2}\end{array}$

Now,

$\mathrm{P}(\mathrm{W}\cap \mathrm{E})=\mathrm{P}(\mathrm{W}\cap {\mathrm{u}}_2)+\mathrm{P}(\mathrm{W}\cap {\mathrm{u}}_4)+...........+\mathrm{P}(\mathrm{W}\cap {\mathrm{u}}_{\mathrm{n}})$

$$\begin{gathered} \Rightarrow \mathrm{P}(\mathrm{W}\cap \mathrm{E})=\frac{1}{\mathrm{n}}\times \frac{2}{\mathrm{n}+1}+\frac{1}{\mathrm{n}}\times \frac{4}{\mathrm{n}+1}+............+\frac{1}{\mathrm{n}}\times \frac{\mathrm{n}}{\mathrm{n}+1} \\ \Rightarrow \mathrm{P}(\mathrm{W}\cap \mathrm{E})=\frac{(\mathrm{n}+2)}{4(\mathrm{n}+1)} \end{gathered}$$

Since, $\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)$ is given by,

$\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}$

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}}\right)=\frac{\mathrm{P}(\mathrm{W}\cap \mathrm{E})}{\mathrm{P}\left(\mathrm{E}\right)}\\ & \Rightarrow & \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}}\right)=\frac{{\displaystyle \frac{(\mathrm{n}+2)}{4(\mathrm{n}+1)}}}{{\displaystyle \frac{1}{2}}}\\ & & \\ & \Rightarrow & \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}}\right)=\frac{(\mathrm{n}+2)}{2(\mathrm{n}+1)}\end{array}$

Hence, option $\left(\mathrm{B}\right)$, $\frac{\mathrm{n}+2}{2(\mathrm{n}+1)}$ is the correct answer.