2001 AIME II Problems/Problem 11
Contents
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 1
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 .
Solution 2
At first, it wins games, only one way
Secondly, it wins games, the other game can be either win or loss, there are ways
Thirdly, it wins games, still the other two games can be either win or loss, there are ways
Fourthly, it wins games, this time, it can't lose games but other arrangements of the three non-winning games are fine, there are ways
Fifth case, it wins games, only lose and draw is ok, so there are cases
Last case, it only wins game so the rest games must be all draw, game
The answer is leads to
~bluesoul
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.