A bag contains `a` white and `b` black balls. Two players, `Aa n dB` 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 find the ratio `a : b`

Answer

`a:b=2:1`

Explanation

Let
W: evetn of drawing a white ball at any draw, and
B: evetn of drawing a black ball at any draw Then
`P(W)=(a)/(a+b)and P(B)=(b)/(a+b)`
`P(A "wins the game")=P(Wor BBWor BBBBW or...)`
`=P(W)+P(B)P(B)P(W)+P(B)P(B)P(B)P(B)P(W)+...`
`=P(W)+P(W)(P(B))^(2)+P(W)P(B))^(4)+....`
`=(P(W))/(1-(P(B))^(2))=(a(a+b))/(a^(2)+2ab)=(a+b)/(a+2b)`
Also, P(B wins the game) `=1-(a+b)/(a+2b)=(a)/(a+2b)`
According to the given condition,
`implies(a+b)/(a+3b)=3xx(b)/(a+2b)`
`impliesa=2b`
`impliesa:b = 2:1`