Difference between revisions of "1996 AIME Problems/Problem 6"
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== Solution == | == Solution == | ||
− | {{ | + | We can use [[complementary counting]]: finding the probability that at least one team wins all games or at least one team loses all games. |
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+ | No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games. | ||
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+ | Now we use [[PIE]]: | ||
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+ | The probability that one team wins all games is <math>\frac{5}{16}</math>. The probability that one team loses all games is <math>\frac{5}{16}</math>. The probability that one team wins all games and another team loses all games is <math>\frac{5}{32}</math>. <math>\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}</math> | ||
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+ | Since that's opposite the probability we want, we subtract that from 1 to get <math>\frac{17}{32}</math>. <math>17+32=\boxed{049}</math> | ||
== See also == | == See also == |
Revision as of 13:59, 12 November 2007
Problem
In a five-team tournament, each team plays one game with every other team. Each team has a chance of winning any game it plays. (There are no ties.) Let
be the probability that the tournament will product neither an undefeated team nor a winless team, where
and
are relatively prime integers. Find
.
Solution
We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games.
No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.
Now we use PIE:
The probability that one team wins all games is . The probability that one team loses all games is
. The probability that one team wins all games and another team loses all games is
.
Since that's opposite the probability we want, we subtract that from 1 to get .
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |