A bag contains $a$ white and $b$ black balls. Two players $A$ and $B$ alternately draw a ball from the bag replacing the ball each time after the draw till one of them draws a white ball and wins the game. $A$ begins the game. If the probability of $A$ winning the game is three times that of $B$, then the ratio $a:b$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Let $W$ denotes the events of drawing a white ball at any draw and $B$ that for a black ball.
$Then\ P(W) = \frac{a}{a + b},\ P(\ B) = \frac{b}{a + b}$
$P(A\ wins\ the\ game) = P(W\ or\ BBBBW\ or\ldots.)$
$= P(W) + P(B)P(B)P(W) + P(B)P(B)P(B)P(B)P(W) + .\ldots$
$= \frac{P(W)}{1 - {P(B)}^{2}} = \frac{\frac{a}{a + b}}{1 - \frac{b^{2}}{(a + b)^{2}}}\frac{a(a + b)}{a^{2} + 2ab} = \frac{(a + b)}{a + 2b}$
$Also\ P(B\ wins\ the\ game) = 1 - \frac{a + b}{a + 2b} = \frac{b}{a + 2b}$
According to the given condition,
$\frac{a + b}{a + 2b} = 3 \bullet \frac{b}{a + 2b}\  \Rightarrow a = 2b\  \Rightarrow a\ :b = 2\ :1$

Explanation

Reviewed by Dr. Abubacker

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