Difference between revisions of "2005 AMC 10A Problems/Problem 18"
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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. | 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 <math>1</math> possibility where team B wins the first game and <math>4</math> total possibilities when team A wins the tournament and team B wins the second game. | + | There is <math>1</math> possibility where team B wins the first game and <math>4</math> total possibilities when team A wins the tournament and team B wins the second game. Note that the fourth possibility (ABAAX) occurs just as often as the others, so we put <math>1</math> over <math>4</math> total possibilities. The desired probability is then <math>\frac{1}{4}\Rightarrow \boxed{B}.</math> |
| − | + | 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 <math>\frac{1}{4}\Rightarrow \boxed{B}.</math> -Flames | We have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is <math>\frac{1}{4}\Rightarrow \boxed{B}.</math> -Flames | ||
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| + | edited by mobius247 | ||
==See Also== | ==See Also== | ||
Revision as of 09:32, 31 May 2021
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.
There is 
possibility where team B wins the first game and 
total possibilities when team A wins the tournament and team B wins the second game. Note that the fourth possibility (ABAAX) occurs just as often as the others, so we put over total possibilities. The desired probability is then 
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 -Flames
edited by mobius247
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. 
