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 $W,G$ denote event of getting a white and a green ball respectively. Then which of the following is/are correct?

• A. If $P\left(U_i\right)\propto i^2,$ then $\underset{n\to \infty }{lim}P\left(G\right)=\frac{1}{4}$
• B. If $P\left(U_i\right)\propto i^2,$ then $\underset{n\to \infty }{lim}P\left(G\right)=\frac{1}{2}$
• C. If $P\left(U_i\right)=k;$ where $k$ is a constant, then $P\left(W\right)=\frac{n+1}{3n+1}$
• D. If $P\left(U_i\right)=k;$ where $k$ is a constant, then $P\left(W\right)=\frac{n+1}{2(3n+1)}$

Answer 1

Answer: (A), (D)

The correct options are
A If $P\left(U_i\right)\propto i^2,$ then $\underset{n\to \infty }{lim}P\left(G\right)=\frac{1}{4}$
D If $P\left(U_i\right)=k;$ where $k$ is a constant, then $P\left(W\right)=\frac{n+1}{2(3n+1)}$

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

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


Now, $\sum P\left(U_i\right)=k$
$\Rightarrow nk=1\Rightarrow k=\frac{1}{n}$

$P\left(W\right)=\frac{1}{n}\times \left(\frac{1}{3n+1}\right)+\frac{1}{n}\times \frac{2}{(3n+1)}+....+\frac{1}{n}\times \frac{n}{(3n+1)}=\frac{1}{n(3n+1)}\left\{\frac{n(n+1)}{2}\right\}=\frac{n+1}{2(3n+1)}$