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

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

Answer 1

The correct option is C

$2:1$

Explanation for the correct option:

Step- 1: Find the ratio of white and black balls:

Let $P\left(A\right)$ be the probability of $A$ wins the game and $P\left(B\right)$ be that of $B$ wins the game

Also, W be the event of drawing a white ball and B be that of drawing a black ball.

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

$\therefore P\left(A\right)=P(WorBBWorBBBBWor......)$

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

$=P\left(W\right)(1+(P{\left(B\right))}^2+(P{\left(B\right))}^4+.....)$

$=\frac{P\left(W\right)}{1-(P{\left(B\right))}^2}$ $\left[\because S_{\infty }=\frac{a}{1-r}\right]$

Step- 2: Substitute the value of $P\left(W\right)$ and $P\left(B\right)$:

$P\left(A\right)=\frac{{\displaystyle \frac{a}{a+b}}}{1-{\displaystyle \frac{b^2}{{(a+b)}^2}}}$

$=\frac{a(a+b)}{a^2+2ab}$

$=\frac{a+b}{a+2b}$

$\because P\left(B\right)=1–P\left(A\right)$

$=1-\frac{(a+b)}{(a+2b)}$

$=\frac{b}{a+2b}$

Step- 3: According to the given condition, $P\left(A\right)=3P\left(B\right)$:

$\Rightarrow$$\frac{a+b}{a+2b}=3\left(\frac{b}{a+2b}\right)$

$\Rightarrow$ $a=2b$

$\Rightarrow$ $a:b=2:1$

Hence, option ‘C’ is correct.