Question

A bag contains $n$ black and 2 red balls. Two persons $A$ and $B$ in order draw one ball from the bag without replacement. A wins as soon as two black balls are drawn and $B$ wins as soon as two red balls are drawn. The game continues until one of the two wins. Let $P(n)$ be the probability that $A$ wins. Then $\underset{n\to \infty }{\mathrm{Lim}}P(n)$ is equal to

• A. $\frac{1}{10}$
• B. $\frac{1}{5}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (D)

Probability that $B$ wins $=\frac{2}{n+2}\cdot \frac{1}{n+1}+\frac{2}{n+2}\cdot \frac{n}{n+1}\cdot \frac{1}{n}+\frac{n}{n+2}\cdot \frac{2}{n+1}\cdot \frac{1}{n}$
$=\frac{6}{(n+1)(n+2)}$
$\therefore \text{ Prob. that A wins }=1-\frac{6}{(n+1)(n+2)}$
$\underset{n\to \infty }{\mathrm{Lim}}P(n)=1$