2005 AMC 10A Problems/Problem 18


Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

$\mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ }  \frac{1}{4}\qquad \mathrm{(C) \ }  \frac{1}{3}\qquad \mathrm{(D) \ }  \frac{1}{2}\qquad \mathrm{(E) \ }  \frac{2}{3}$


There are at most $5$ games played.

If team B won the first two games, team A would need to win the next three games. So the only possible order of wins is BBAAA.

If team A won the first game, and team B won the second game, the possible order of wins are: ABBAA, ABABA, and ABAAX, where X denotes that the 5th game wasn't played.

Since ABAAX is dependent on the outcome of $4$ games instead of $5$, it is twice as likely to occur and can be treated as two possibilities. (If $5$ games were played, X could be A or B)

Since there is $1$ possibility where team B wins the first game and $5$ total possibilities, the desired probability is $\frac{1}{5}\Rightarrow \boxed{A}$

Video Solution

CHECK OUT Video Solution: https://youtu.be/xqdc2N6fckA

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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