Question

There are $n$ urns each containing $3n+1$ balls such that $i^{\text{th}}$ urn contains $i$ white, $(i+1)$ green and $(3n–2i)$ red balls. Let $U_i$ be the event of selecting $i^{\text{th}}$ urn, where $i=1,2,3,...,n$ and $G$ denotes the event of getting a green ball. If If $P\left(U_i\right)\propto i^2,$ then $\underset{n\to \infty }{lim}P\left(G\right)$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{1}{4}$

If $P\left(U_i\right)\propto i^2,$ then
$P\left(U_i\right)=k{i}^2,$ where $k$ is proportionality constant.
$\therefore P\left(U_1\right)+P\left(U_2\right)+...+P\left(U_n\right)=1\Rightarrow k\times 1^2+k\times 2^2+k\times 3^2+...+k\times n^2=1\Rightarrow k\times \frac{n(n+1)(2n+1)}{6}=1\Rightarrow k=\frac{6}{n(n+1)(2n+1)}$

$\underset{n\to \infty }{lim}P\left(G\right)=\underset{n\to \infty }{lim}\left[P\right(U_1)P\left(\frac{G}{U_1}\right)+P(U_2)P\left(\frac{G}{U_2}\right)+...+P(U_n\left)P\left(\frac{G}{U_n}\right)\right]=\underset{n\to \infty }{lim}\sum _{i=1}^{n}k\frac{i^2(i+1)}{(3n+1)}=\underset{n\to \infty }{lim}\frac{6[\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{6}]}{n(n+1)(2n+1)(3n+1)}=\frac{6}{4\cdot 2\cdot 3}=\frac{1}{4}$