Question

There are $n$ urns, each of these contain $n+1$ balls. The $i^{th}$ urn contains $i$ white balls and $n+1-i$ red balls. Let $u_i$ be the event of selecting $i^{th}$ urn, $i=1,2,3,\cdots n$ and $w$ be the event of getting a white ball.
If $P\left(u_i\right)\propto i$, where $i=1,2,3\cdots n,$ then $\underset{n\to \infty }{lim}P\left(w\right)$ equals to

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

Answer 1

The correct option is B $\frac{2}{3}$

Let $P\left(u_i\right)=Ki$ , $K$ being proportionality constant .
Then $\sum _{i=1}^{n}P\left(u_i\right)=1\Rightarrow (\sum Ki)=1\Rightarrow K=\frac{2}{n(n+1)}$
Each urn contains $(n+1)$ balls and $i^{th}$ urn contains $i$ white balls.

$\therefore P\left(w\right)=\sum p\left(u_i\right)\times \left(\frac{i}{n+1}\right)=\sum \frac{K{i}^2}{(n+1)}=\frac{2}{n(n+1{)}^2}\left[\frac{n(n+1)(2n+1)}{6}\right]$
$\underset{n\to \infty }{lim}P\left(w\right)=\left[\frac{\text{coefficient of}{n}^3\text{in}Nr}{\text{coefficient of}{n}^3\text{in}Dr}\right]=\frac{4}{6}=\frac{2}{3}$