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 $\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)=\mathrm{c}$, where $\mathrm{c}$ is a constant, then $\mathrm{P}\left(\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{W}}\right)$ is equal to

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Evaluate the required value based on the given information

Given, $\mathrm{W}$ denotes the event of getting a white ball.

Probability of selecting $i^{th}$urn when drawn ball is white is given as $P\left(\frac{u_n}{W}\right)$

Thus, $\mathrm{P}\left(\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{W}}\right)=\frac{\left({\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}}\right)}{\left({\displaystyle \frac{{\displaystyle \sum _{}}\mathrm{i}}{\mathrm{n}+1}}\right)}$

$=\frac{{\displaystyle \frac{n}{n+1}}}{{\displaystyle \frac{n(n+1)}{2(n+1)}}}$

$\Rightarrow$ $\mathrm{P}\left(\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{W}}\right)=\frac{2}{\left(\mathrm{n}+1\right)}$

Hence option (A) is the correct option.