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