Question

There are n urns $u_1,u_2,\dots \dots u_n$. Each urn contains $\left(3^{n+1}\right)$ balls. The $i^{th}$ urn contains 3i number of white balls. $P\left(u_i\right)$, i.e. Probability of selecting $i^{th}$ urn is proportional to $(i^2+3)$. If we randomly select one of the urns and draw one ball and probability of ball being white be P(A), then let $\underset{n\to \infty }{\text{lim}}P\left(A\right)$ is 'a/b'. Value of a+b is

Answer 1

$P\left(u_i\right)\propto (i^2+3)$
$\Rightarrow P\left(u_i\right)=k(i^2+3)$
$u_1,u_2,\dots \dots u_n$ are mutually exclusive events.
$\therefore k\left[\right(1^2+3)+\dots \dots +(n^2+3\left)\right]=1$
$k=\frac{1}{\frac{n(n+1)(2n+1)}{6}+3n}=\frac{6}{n(n+1)(2n+1)+18n}$
So $P\left(A\right)=k(1^2+3)\times \frac{3}{3^{n+1}}+k(2^2+3)\times \frac{6}{3^{n+1}}+\dots \dots k(n^2+3)\times \frac{3n}{\left(3^{n+1}\right)}$
$=\frac{3k}{(3n+1)}\left[\right(1^2+3)+2.(2^2+3)+\dots \dots +n(n^2+3\left)\right]$
$=\frac{3k}{(3n+1)}\left[\right(1^3+2^3+\dots \dots +n^3)+3(1+2+\dots \dots +n\left)\right]$
$=\frac{3}{(3n+1)}\times \frac{6}{\text{/}n\left[\right(n+1\left)\right(2n+1)+18]}\times \left(\frac{\text{/}n(n+1)}{4}\left(n\right(n+1)+6)\right)$
$=\frac{9}{2}\frac{\left(n\right(n+1{)}^2+6(n+1))}{\left[\right(n+1\left)\right(2n+1\left)\right(3n+1)+18(3n+1\left)\right]}$
So, $\underset{n\to \infty }{\text{lim}}P\left(A\right)=\frac{9}{2}\times \frac{1}{2\times 3}=\frac{3}{4}=0.75$