Difference between revisions of "2001 AIME II Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each <math>\frac {1}{3}</math>. The probability that Club Truncator will finish the season with more wins than losses is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | + | Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each <math>\frac {1}{3}</math>. The [[probability]] that Club Truncator will finish the season with more wins than losses is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. |
== Solution == | == Solution == | ||
| − | + | Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the [[complement principle]], the desired probability is half the probability that Club Truncator does not have the same number of wins and losses. | |
| + | |||
| + | The possible ways to achieve the same number of wins and losses are <math>0</math> ties, <math>3</math> wins and <math>3</math> losses; <math>2</math> ties, <math>2</math> wins, and <math>2</math> losses; <math>4</math> ties, <math>1</math> win, and <math>1</math> loss; or <math>6</math> ties. Since there are <math>6</math> games, there are <math>\frac{6!}{3!3!}</math> ways for the first, and <math>\frac{6!}{2!2!2!}</math>, <math>\frac{6!}{4!}</math>, and <math>1</math> ways for the rest, respectively, out of a total of <math>3^6</math>. This gives a probability of <math>141/729</math>. Then the desired answer is <math>\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}</math>, so the answer is <math>m+n = \boxed{341}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=10|num-a=12}} | {{AIME box|year=2001|n=II|num-b=10|num-a=12}} | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 09:00, 27 July 2008
Problem
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each 
. The probability that Club Truncator will finish the season with more wins than losses is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle, the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
The possible ways to achieve the same number of wins and losses are 
ties, 
wins and losses; 
ties, wins, and losses; 
ties, 
win, and loss; or 
ties. Since there are games, there are 
ways for the first, and 
, 
, and ways for the rest, respectively, out of a total of 
. This gives a probability of 
. Then the desired answer is 
, so the answer is 
.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
