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. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.
Let’s take A1 to be the event of getting a total of 6
A1 = {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)}
And, B1 be the event of getting a total of 7
B1 = {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)}
Let P(A1) is the probability, if A wins in a throw = 5/36
And P(B1) is the probability, if B wins in a throw = 1/6
Therefore, the required probability of wining A in his third throw
$=P\left(\bar{A}_{1}\right) \cdot P\left(\bar{B}_{1}\right) \cdot P\left(A_{1}\right)=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{7776}$