Difference between revisions of "2005 AMC 10A Problems/Problem 18"
(typo in solution) |
|||
| Line 23: | Line 23: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 11:30, 4 July 2013
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?
Solution
There are at most 
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 
games instead of , it is twice as likely to occur and can be treated as two possibilities.
Since there is 
possibility where team B wins the first game and total possibilities, the desired probability is 
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
