Question

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 $B$, the ratio $a:b$ is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. None of these

Answer 1

Answer: (C)

$2:1$

Let $W$ denote the event of drawing a white ball at any draw and $B$ that for a black ball. Then

$P\left(W\right)=\frac{a}{a+b}$ and $P\left(B\right)=\frac{b}{a+b}$

$P(A$ wins the game$)=P(W$ or $BBW$ or $BBBBW$ or $...)$

$=P\left(W\right)+P\left(BBW\right)+P\left(BBBBW\right)+...$

$=P\left(W\right)+P\left(B\right)P\left(B\right)P\left(W\right)+P\left(B\right)P\left(B\right)P\left(B\right)P\left(B\right)P\left(W\right)+...$

$=P\left(W\right)+P\left(W\right).P{\left(B\right)}^2+P\left(W\right).P{\left(B\right)}^4+...$

$=\frac{P\left(W\right)}{1-P{\left(B\right)}^2}=\frac{a(a+b)}{a^2+2ab}=\frac{a+b}{a+2b}$

Also $P(B$ wins the game $)=1-\frac{a+b}{a+2b}=\frac{b}{a+2b}$

According to the given conditions,

$\Rightarrow \frac{a+b}{a+2b}=3.\frac{b}{a+2b}\Rightarrow a=2b\Rightarrow a:b=2:1$