There are n urns each containing $n+1$ balls such that the ith urn contains i white balls and $(n+1-i)$ red balls. Let $u_{i}$ be the event of selecting ith urn, $i=1,2,3 \ldots, n$ and w denotes the event of getting a white ball. If $P(u_{i}) \propto i$, where $i=1,2,3, \ldots n$, then $\lim_{n \rightarrow \infty} P(w)$ is equal to

JEE Advanced · 2006 · 09 Apr 2006 · Probability XII

There are n urns each containing

n + 1

balls such that the ith urn contains i white balls and

(n + 1 - i)

red balls. Let

u_{i}

be the event of selecting ith urn,

i = 1, 2, 3 \dots, n

and w denotes the event of getting a white ball.

If

P (u_{i}) \propto i

, where

i = 1, 2, 3, \dots n

, then

lim_{n \to \infty} P (w)

is equal to

Answer: (B)

Given

P (u_{i}) \propto i

, let

P (u_{i}) = k i

for some constant

k

.

Since

\sum_{i = 1}^{n} P (u_{i}) = 1

k i = 1 \sum n i = 1 \Rightarrow k \cdot \frac{n ( n + 1 )}{2} = 1 \Rightarrow k = \frac{2}{n ( n + 1 )}

The probability of drawing a white ball from urn

i

\frac{i}{n + 1}

.

By the law of total probability:

P (w) = i = 1 \sum n P (u_{i}) \cdot P (w ∣ u_{i}) = i = 1 \sum n \frac{2 i}{n ( n + 1 )} \cdot \frac{i}{n + 1} = \frac{2}{n ( n + 1 ) ^{2}} i = 1 \sum n i^{2}

Using

\sum_{i = 1}^{n} i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

P (w) = \frac{2}{n ( n + 1 ) ^{2}} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 ( 2 n + 1 )}{6 ( n + 1 )} = \frac{2 n + 1}{3 ( n + 1 )}

Taking the limit:

n \to \infty lim P (w) = n \to \infty lim \frac{2 n + 1}{3 n + 3} = \frac{2}{3}

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option