Question

A bag contains a while balls 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 the ball is white and win the game. The player A begins the game. If the probability of A winning the game is four times that of B, then the ratio a : b is

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

Answer 1

The correct option is B 3 : 1

Let W – white, B – black ball
$P\left(W\right)=\frac{a}{a+b},P\left(B\right)=\frac{b}{a+b}$
P(A wins the game) = P(W) + P(BBW) + p(BBBBW) + . . . ..
= P(W) + P(B) ∙ P(B) ∙ P(W) + . . . . .
$=P\left(W\right)(1+P(B{)}^2+P(B{)}^4+.....)$
$=\frac{P\left(W\right)}{1-P(B{)}^2}=\frac{a+b}{a+2b}$
P(B wins the game) = 1 – P(A wins the game) = $1-\frac{a+b}{a+2b}=\frac{b}{a+2b}$
Given that $\frac{a+b}{a+2b}=4\frac{b}{a+2b}\Rightarrow a=3b$
$\therefore a:b=3:1$