- #1

roflmao33

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## Homework Statement

Two teams, teams A and B, are competing in a tournament. Suppose the tournament is a ‘best-of-3’ series; in other words, the first team to win 2 games wins the tournament. Let’s suppose that team A has a slight edge on team B, such that the probability that team A wins any given game is 60%, and the outcome of any previous match does not influence team A’s probability of winning in any future match.

a. What is the probability that team A wins this tournament?

b. Explain why the probability that team A wins the tournament is greater than 60%.

c. If this had been a ‘best-of-7’ series (first team to win 4 games), would the probability that team A wins the tournament be greater or less than what you calculated in a)?

d. Calculate the probability team A wins a ‘best-of-7’ series.

## The Attempt at a Solution

There should be 3 ways that team A can win the tournament (3 choose 2=3). Since the probability of team A winning a game is 0.6. The probability of team A winning the tournament will be 3C2*(0.6)^2*(0.4)=0.432?

What am I doing wrong. I can't get a probability of greater than 60% as of question (b).

Thx in advance.