$A$ and $B$ throw a pair of dice alternately. $A$ wins the game, if he gets a total of 6 and $B$ wins, if she gets a total of 7. If $A$ starts the game, find the probability of winning the game by $A$ in third throw of the pair of dice.

$\frac{775}{7776}$

The correct answer is Option (2) → $\frac{775}{7776}$ ##

Let $A_1 = \text{A total of } 6 = \{(2, 4), (1, 5), (5, 1), (4, 2), (3, 3)\}$

and $B_1 = \text{A total of } 7 = \{(2, 5), (1, 6), (6, 1), (5, 2), (3, 4), (4, 3)\}$

Let $P(A_1)$ is the probability, if $A$ wins in a throw $\Rightarrow P(A_1) = \frac{5}{36}$

$∴P(\overline{A_1}) = 1 - P(A_1) = 1 - \frac{5}{36} = \frac{31}{36}$

and $P(B_1)$ is the probability, if $B$ wins in a throw $\Rightarrow P(B_1) = \frac{1}{6}$

$∴P(\overline{B_1}) = 1 - P(B_1) = 1 - \frac{1}{6} = \frac{5}{6}$

$∴$ Required probability $= P(\overline{A_1}) \cdot P(\overline{B_1}) \cdot P(A_1) = \frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36} = \frac{775}{7776}$