@engineer,
I'd agree with engineer if the question were, "What is the probability that a team (either A or B) wins the series in exactly 6 games, and one team (either A or B) won both the first and second game?"
However, we are
given that one team has already won the first two games. This eliminates all six-game series where the first two games are split.
A wins the first two games with probability 0.68^2 = 0.4624
B wins the first two games with probability 0.32^2 = 0.1024
Possible series and final four-game probabilities when A wins first two games:
AA ABBA 0.68^2 * 0.32^2
AA BABA 0.68^2 * 0.32^2
AA BBAA 0.68^2 * 0.32^2
AA BBBB 0.32^4
AA total is 3 * 0.68^2 * 0.32^2 + 0.32^4 = 0.15253504
Possible series and final four-game probabilities when B wins first two games:
BB BAAB 0.68^2 * 0.32^2
BB ABAB 0.68^2 * 0.32^2
BB AABB 0.68^2 * 0.32^2
BB AAAA 0.68^4
BB total is 3 * 0.68^2 * 0.32^2 + 0.68^4 = 0.35586304
The totals need to be weighted based on the relative probabilities of AA and BB in the first two games. Therefore the grand total is:
0.15253504 * [0.4624 / (0.4624 + 0.1024)] + 0.35586304 * [0.1024 / (0.4624 + 0.1024)] = 0.18939904