A bag contains a white and b black balls. Two | vedclass

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Question

(C) Let $W$ denote the event of drawing a white ball and $B$ denote the event of drawing a black ball.The probability of drawing a white ball is $P(W)

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(C) Let $W$ denote the event of drawing a white ball and $B$ denote the event of drawing a black ball.The probability of drawing a white ball is $P(W) = \frac{a}{a+b}$ and the probability of drawing a black ball is $P(B) = \frac{b}{a+b}$.Player $A$ wins if they draw a white ball on their turn ($1^{st}, 3^{rd}, 5^{th}, \dots$ turns).$P(A 	ext{ wins}) = P(W) + P(B)P(B)P(W) + P(B)P(B)P(B)P(B)P(W) + \dots$This is a geometric series with first term $a_1 = P(W) = \frac{a}{a+b}$ and common ratio $r = P(B)^2 = \frac{b^2}{(a+b)^2}$.$P(A 	ext{ wins}) = \frac{P(W)}{1 - P(B)^2} = \frac{\frac{a}{a+b}}{1 - \frac{b^2}{(a+b)^2}} = \frac{a(a+b)}{(a+b)^2 - b^2} = \frac{a(a+b)}{a^2 + 2ab} = \frac{a+b}{a+2b}$.Since $P(A 	ext{ wins}) + P(B 	ext{ wins}) = 1$,we have $P(B 	ext{ wins}) = 1 - \frac{a+b}{a+2b} = \frac{b}{a+2b}$.Given that $P(A 	ext{ wins}) = 3 	imes P(B 	ext{ wins})$:$\frac{a+b}{a+2b} = 3 	imes \frac{b}{a+2b}$.$a+b = 3b \implies a = 2b \implies \frac{a}{b} = \frac{2}{1}$.Thus,the ratio $a:b$ is $2:1$.

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