Difference between revisions of "2005 AMC 10A Problems/Problem 18"
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==Problem== | ==Problem== | ||
| − | Team A and team B play a series. The first team to win three games wins the series. | + | 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? |
| − | <math> \ | + | <math> \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} </math> |
==Solution== | ==Solution== | ||
There are at most <math>5</math> games played. | There are at most <math>5</math> 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 <math>B</math> won the first two games, team <math>A</math> would need to win the next three games. So the only possible order of wins is <math>BBAAA</math>. |
| − | 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 | + | If team <math>A</math> won the first game, and team <math>B</math> won the second game, the possible order of wins are: <math>ABBAA, ABABA,</math> and <math>ABAAX</math>, where <math>X</math> denotes that the <math>5</math>th game wasn't played. |
| − | + | There is <math>1</math> possibility where team <math>B</math> wins the first game and <math>4</math> total possibilities when team <math>A</math> wins the series and team <math>B</math> wins the second game. Note that the fourth possibility <math>(ABAAX)</math> occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities. The desired probability is then <math>\boxed{\textbf{(A) }\frac{1}{5}}</math>. | |
| − | + | ==Note== | |
| + | The original final problem was poorly worded, since the problem directly stated that the answer is <math>\boxed{1/2}</math>. | ||
| + | |||
| + | The problem should say "what fraction of possible sets of game outcomes have <math>B</math> winning the first game?" or "Given the observed results, what is the conditional probability that <math>B</math> won the first game?" | ||
| + | |||
| + | (Many problems in probability are poorly worded.) | ||
==See Also== | ==See Also== | ||
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| − | + | {{AMC10 box|year=2005|ab=A|num-b=17|num-a=19}} | |
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:55, 9 July 2024
Contents
Problem
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?
Solution
There are at most 
games played.
If team 
won the first two games, team 
would need to win the next three games. So the only possible order of wins is 
.
If team won the first game, and team won the second game, the possible order of wins are: 
and 
, where 
denotes that the th game wasn't played.
There is 
possibility where team wins the first game and 
total possibilities when team wins the series and team wins the second game. Note that the fourth possibility 
occurs twice as often as the others, so we put over total possibilities. The desired probability is then 
.
Note
The original final problem was poorly worded, since the problem directly stated that the answer is 
.
The problem should say "what fraction of possible sets of game outcomes have winning the first game?" or "Given the observed results, what is the conditional probability that won the first game?"
(Many problems in probability are poorly worded.)
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. 
