Question

A bag contains $a$ white and $b$ black balls. Two players $A$ and $B$ alternatively 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 As winning the game is three times that of $B$, then the ratio $a:b$ is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:3$

Answer 1

The correct option is C $2:1$

the probability that $A$ wins is:
$(a/a+b)+(b/a+b)\ast (b/a+b)\ast (a/a+b)+..........$till infinity
taking out $a/a+b$ common and then applying the formula of infinite gp series we have
$P\left(A\right)=a+b/a+2b$
Now, finding the probability of $B$
$(b/a+b)\ast (a/a+b)+(b/a+b)\ast (b/a+b)\ast (b/a+b)\ast (a/a+b)+...........$till infinity
Taking out $(b/a+b)\ast (a/a+b)$ common and applying the formula of infinite gp we have
P(B)=$\frac{b}{a}+2b$
NOW in the question it has been clearly mentioned that $P\left(A\right)=3P\left(B\right)$
...therefore by solving it we are getting the ratio to be $2:1$(ANS)