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 n is even and E denotes the event of choosing even numbered urn $(P(u_{i})=\frac{1}{n})$, then the value of $P(w/E)$ is

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 n is even and E denotes the event of choosing even numbered urn

(P (u_{i}) = \frac{1}{n})

, then the value of

P (w / E)

Answer: (B)

Given

n

is even and

E

is the event of choosing an even-numbered urn.

The even-numbered urns are

2, 4, 6, \dots, n

(there are

\frac{n}{2}

such urns).

Given

E

, each even urn is equally likely:

P (u_{i} ∣ E) = \frac{2}{n}

for even

i

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

= \frac{2}{n ( n + 1 )} i even \sum i = \frac{2}{n ( n + 1 )} (2 + 4 + 6 + \dots + n)

= \frac{2}{n ( n + 1 )} \cdot 2 (1 + 2 + 3 + \dots + \frac{n}{2}) = \frac{2}{n ( n + 1 )} \cdot 2 \cdot \frac{\frac{n}{2} ( \frac{n}{2} + 1 )}{2}

= \frac{2}{n ( n + 1 )} \cdot \frac{n ( \frac{n}{2} + 1 )}{2} = \frac{\frac{n}{2} + 1}{n + 1} = \frac{n + 2}{2 ( n + 1 )}

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