Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. The probability of A's winning if A starts first is

$\frac{6}{11}$

The correct answer is Option (1) → $\frac{6}{11}$ **

Let probability of getting a 3 on a single throw = $\frac{1}{6}$, and probability of not getting a 3 = $\frac{5}{6}$.

Let $P$ = probability that A wins the game.

A can win in the 1st, 3rd, 5th, ... turns.

Hence, $P = \frac{1}{6} + \left(\frac{5}{6}\right)^2\frac{1}{6} + \left(\frac{5}{6}\right)^4\frac{1}{6} + \dots$

This is an infinite geometric series with first term $a = \frac{1}{6}$ and common ratio $r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

$P = \frac{a}{1 - r} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{36}{66} = \frac{6}{11}$

Final Answer:

$P = \frac{6}{11}$