Hello MATHMATHJ28,
We split this problem by tackling three cases:
CASE 1:
Team A wins all three games.
This comes with probability (2/3)^3 = 8/27
CASE 2:
Team A wins 3 games, and Team B wins one game
This has probability: ((2/3)^3)*(1/3)*3
There are three ways to arrange 2 games won by A and one by B. The last game will always be won by A. So, there are 3 ways to arrange this.
This is why we multiply 3 at the end. Finally, we get 8/27.
CASE 3:
Team A wins 3 games, and Team B wins 2 games.
Using the same strategy as before, we know Team A must always win the last game. So we count the number of ways for Team A to achieve 2 wins ans Team B to achieve two wins. This can be done in 4 choose 2 = 6 ways. We get:
((2/3)^3)*((1/3)^2)*(6)
=16/81
If we add all of these fractions up we get our answer:
8/27 + 8/27 + 16/81
64/81
Saffyre:)
