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# Team A is playing against Team B in a soccer tournament. The first team to win 3 games wins the tournament.

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Team A is playing against Team B in a soccer tournament. The first team to win 3 games wins the tournament. Team A has a probability of 2/3 of winning each game. What is the probability that Team A wins the tournament?

Mar 15, 2020

#1
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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:)

Mar 15, 2020
#3
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Thanks so much for helping me understand the problem!!

mathmathj28  Mar 15, 2020
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