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 $P(u_{i})=c$, where c is a constant then $P(u_{n}/w)$ 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}) = c

, where c is a constant then

P (u_{n} / w)

is equal to

Answer: (A)

Given

P (u_{i}) = c

for all

i

, and since

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

n c = 1 \Rightarrow c = \frac{1}{n}

By Bayes' theorem:

P (u_{n} ∣ w) = \frac{P ( w ∣ u _{n} ) \cdot P ( u _{n} )}{P ( w )}

The probability of white ball from urn

n

P (w ∣ u_{n}) = \frac{n}{n + 1}

.

The total probability of white ball:

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

Therefore:

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

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