2001 AIME II Problems/Problem 11
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
We first find out how many way there are for the Truncators to win and lose the same amount. They can get 0 ties, 3 wins and 3 losses, 2 ties, 2 wins, and 2 losses, 4 ties, 1 win, and 1 loss, or 6 ties. Since there are 6 games, there are 6!/3!3! ways for the first, and 6!/2!2!2!, 6!/4!, and 1 ways for the rest, respectively, out of a total of 
. This gives a probability of 141/729. 1 minus this is the probability they won't win and lose the same number of games, and half that is the probability they'll finish with more wins. This is 98/243, so the answer is 341.
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 | ||
