Suppose that two chess players had played numerous games and

Suppose that two chess players had played numerous games and it was determined that the probability that Player A would win is 0.40, the probability that Player B would win is 0.35, and the probability that the game would end in a draw is 0.25. If these two chess players played 12 games, what is the probability that Player A would win 7 games, Player B would win 2 games, and the remaining 3 games would be drawn?

Solution

P(A) =0.40 and P(B) = 0.35

P(AB) = 0.25 (i.e. a draw)

OUt of 12 games A win 7 games, B 2 games and 3 games drawn

A can win any 7 out of 12 games.

and B can win any2 out of remaining 5 games

and the remaining 3 draw.

Hence probability = 12C7 (0.40)⁷[5C2(0.35)²](0.25)³

=0.024837