2005 AMC 10A Problems/Problem 18


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

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


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 $5$th 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 twice as often as the others, so we put $1$ over $5$ total possibilities. The desired probability is then $\boxed{\textbf{(A) }\frac{1}{5}}$.


The original final problem was poorly worded, since the problem directly stated that the answer is $\boxed{1/2}$.

The problem should say "what fraction of possible sets of game outcomes have $B$ winning the first game?" or "Given the observed results, what is the conditional probability that $B$ won the first game?"

(Many problems in probability are poorly worded.)

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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