1996 AIME Problems/Problem 6: Difference between revisions
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Now we use [[PIE]]: | Now we use [[PIE]]: | ||
The probability that one team wins all games is <math> | The probability that one team wins all games is <math>5\cdot (\frac{1}{2})^4=\frac{5}{16}</math>. | ||
Since | Similarity, 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>(5\cdot (\frac{1}{2})^4)(4\cdot (\frac{1}{2})^3)=\frac{5}{32}</math>. | |||
<math>\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}</math> | |||
Since this is the opposite of 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 15:47, 16 March 2009
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 
.
Similarity, 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 this is the opposite of 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 | ||
