2005 AMC 10A Problems/Problem 18

Problem

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?

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

Solution

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.

There is $1$ possibility where team B wins the first game and $4$ total possibilities when team A wins the series and team B wins the second game. Note that the fourth possibility (ABAAX) occurs just as often as the others, so we put $1$ over $4$ total possibilities. The desired probability is then $\frac{1}{4}\Rightarrow \boxed{B}.$

The fourth possibility (ABAAX) simply counts as 1 case, and it does not occur twice as often as the others because we are given the information that team A will win the series after winning three games by the problem.

We have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is $\frac{1}{4}\Rightarrow \boxed{B}.$ -Flames

edited by mobius247

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2005 AMC 10A Problems/Problem 18

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