Difference between revisions of "2005 AMC 10A Problems/Problem 18"
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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. However, note that the fourth possibility (ABAAX) occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities instead of <math>4</math>. the desired probability is then <math>\frac{1}{5}\Rightarrow \boxed{A}.</math> | 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. However, note that the fourth possibility (ABAAX) occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities instead of <math>4</math>. the desired probability is then <math>\frac{1}{5}\Rightarrow \boxed{A}.</math> | ||
| − | Note: The solution given to this problem above is incorrect (you can check the answer key of this test to be sure of this). In actuality, the fourth possibility (ABAAX) simply counts as 1 case, and it does not occur twice as often as the others because even though we don't have to multiply by <math>1/2</math> at the end, we are given the information that team A will win by the information in the problem. Thus, we only have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is <math>\frac{1}{4}\Rightarrow \boxed{B}.</math> | + | Note: The solution given to this problem above is incorrect (you can check the answer key of this test to be sure of this). In actuality, the fourth possibility (ABAAX) simply counts as 1 case, and it does not occur twice as often as the others because even though we don't have to multiply by <math>1/2</math> at the end, we are given the information that team A will win by the information in the problem. Thus, we only have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is <math>\frac{1}{4}\Rightarrow \boxed{B}.</math> -Flames |
==See Also== | ==See Also== | ||
Revision as of 02:04, 15 February 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. However, note that the fourth possibility (ABAAX) occurs twice as often as the others, so we put over total possibilities instead of . the desired probability is then 
Note: The solution given to this problem above is incorrect (you can check the answer key of this test to be sure of this). In actuality, the fourth possibility (ABAAX) simply counts as 1 case, and it does not occur twice as often as the others because even though we don't have to multiply by 
at the end, we are given the information that team A will win by the information in the problem. Thus, we only have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is 
-Flames
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. 
