Difference between revisions of "2005 AMC 10A Problems/Problem 18"
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Since there is <math>1</math> possibility where team B wins the first game and <math>4</math> total possibilities, the desired probability is <math>\frac{1}{4}\Rightarrow \boxed{B}</math> | Since there is <math>1</math> possibility where team B wins the first game and <math>4</math> total possibilities, the desired probability is <math>\frac{1}{4}\Rightarrow \boxed{B}</math> | ||
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==See Also== | ==See Also== | ||
Revision as of 15:45, 8 December 2020
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 there is 
possibility where team B wins the first game and 
total possibilities, the desired probability is 
See Also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
