Each of the n bags contains ‘a’ white and ‘b’ black balls. One ball is transferred from 1^st bag to the second bag. Then one ball is transferred from second bag to the third bag and so on. Let $P_{n}$ be the probability that ball transferred from nth bag is white. Then

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$P_{4} = \frac{a}{a + b}$

$P_{1} = \frac{a}{a + b}$

$P_{2} = \frac{a}{a + b}$

$P_{3} = \frac{a}{a + b}$

answer is A, B, C, D.

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Detailed Solution

$p_{1} = \frac{a}{a + b}$
$p_{2} = \frac{a}{a + b} \times \frac{a + 1}{a + b + 1} + \frac{b}{a + b} \times \frac{a}{a + b + 1} = \frac{a}{a + b}$
Let us assume $p_{n} = \frac{a}{a + b}$ for some n then,
$p_{n + 1} = p_{n} \times \frac{a + 1}{a + b + 1} + (1 - p_{n}) \frac{a}{a + b + 1}$
Hence $p_{n + 1} = \frac{a}{a + b} \times \frac{a + 1}{a + b + 1} + \frac{b}{a + b} \times \frac{a}{a + b + 1} = \frac{a}{a + b}$
Hence $p_{n} = \frac{a}{a + b} \forall n$