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 $\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)\propto \mathrm{i}$, where $\mathrm{i}=1,2,3,....\mathrm{n}$, then ${\lim }_{\mathrm{n}\to \infty }$

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B

$\frac{2}{3}$

Evaluate the limit of the given probability

Let $\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)={\mathrm{k}}_{\mathrm{i}}$.

We know that $\sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)=1$.

$\Rightarrow$$\mathrm{k}\left(1+2+3+..+\mathrm{n}\right)=1$

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

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

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{W}\right)& =& \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{P}\left({\mathrm{u}}_{\mathrm{i}}\right)\mathrm{P}\left(\frac{\mathrm{W}}{{\mathrm{u}}_{\mathrm{i}}}\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{W}\right)=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{k}}_{\mathrm{i}}\frac{\mathrm{i}}{\mathrm{n}+1}\\ & \Rightarrow & \mathrm{P}\left(\mathrm{W}\right)=\frac{2}{\mathrm{n}{(\mathrm{n}+1)}^2}\times \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\\ & \Rightarrow & \mathrm{P}\left(\mathrm{W}\right)=\frac{1}{3}\frac{2+{\displaystyle \frac{1}{\mathrm{n}}}}{1+{\displaystyle \frac{1}{\mathrm{n}}}}\end{array}$

Now, evaluate the value of $\underset{\mathrm{n}\to \infty }{\lim }\mathrm{P}\left(\mathrm{W}\right)$.

Thus, $\underset{\mathrm{n}\to \infty }{\lim }\mathrm{P}\left(\mathrm{W}\right)=\frac{1}{3}\times 2$

$\Rightarrow$ $\underset{\mathrm{n}\to \infty }{\lim }\mathrm{P}\left(\mathrm{W}\right)=\frac{2}{3}$

Hence option $\left(\mathrm{B}\right)$, $\frac{2}{3}$ is the correct answer.