A and B throw a die alternatively till one of them gets a '6' and win…

The correct answer is Option (2) → A: $\frac{6}{11}$, B: $\frac{5}{11}$ ##

Let $S$ denote the success (getting a '6') and $F$ denote the failure (not getting a '6').

Thus, $P(S) = \frac{1}{6}, P(F) = \frac{5}{6}$

$P(\text{A wins in the first throw}) = P(S) = \frac{1}{6}$

A gets the third throw, when the first throw by A and second throw by B result into failures.

Therefore, $P(\text{A wins in the 3rd throw}) = P(FFS) = P(F)P(F)P(S) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}$

$= \left(\frac{5}{6}\right)^2 \times \frac{1}{6}$

$P(\text{A wins in the 5th throw}) = P(FFFFS) = \left(\frac{5}{6}\right)^4 \left(\frac{1}{6}\right) \text{ and so on.}$

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

$= \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{6}{11}$

$P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{5}{11}$