Question

A bag contain ${}^{\prime }{a}^{\prime }$ white and ${}^{\prime }{b}^{\prime }$ black ball. Two players $A$ and $B$ alternately draw a ball from the bag, replacing the ball each time after the draw. A begins the game, if the probability of $A$ winning (that is drawing a white ball) is twice the probability of $B$ winning, then the ratio $a:b$ is equal to

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

Answer 1

Answer: (C)

$1:1$

Let $W$ denote the event of drawing ${}^{\prime }{a}^{\prime }$ white ball at any draw

and $B$ denote the event of drawing ${}^{\prime }{b}^{\prime }$ black ball at any draw. Then
$P\left(W\right)=\frac{a}{a+b}$ and $P\left(B\right)=\frac{b}{a+b}$
Now $P(A$ wins the game $)=P(W$ or $BBW$ or $BBBBW$ or $)$
$=P\left(W\right)+P\left(BBW\right)+P\left(BBBBW\right)+...=P\left(W\right)+{\left(P\left(B\right)\right)}^{2}P\left(W\right)+{\left(P\left(B\right)\right)}^{4}P\left(W\right)+...$
$=\frac{P\left(W\right)}{1-{\left(P\left(B\right)\right)}^2}=\frac{\frac{a}{a+b}}{1-\frac{b^2}{{(a+b)}^2}}=\frac{a(a+b)}{{(a+b)}^2-b^2}$
$=\frac{a(a+b)}{a(a+2b)}=\frac{a+b}{a+2b}$
and $P(B$ wins the game $=1-\frac{a+b}{a+2b}=\frac{b}{a+2b}$
According to the given condition
$\frac{a+b}{a+2b}=2.\frac{b}{a+2b}\Rightarrow a=b$
Therefore $a:b::1:1$