Difference between revisions of "2006 AIME II Problems/Problem 10"
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You can break this into cases based on how many rounds A wins out of the remaining 5 games. | You can break this into cases based on how many rounds A wins out of the remaining 5 games. | ||
| − | If A wins 0 games, then B must win 0 games and the probability of this is <math> \frac{{ | + | If A wins 0 games, then B must win 0 games and the probability of this is <math> \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} </math>. |
| − | If A wins 1 games, then B must win 1 or less games and the probability of this is <math> \frac{{ | + | If A wins 1 games, then B must win 1 or less games and the probability of this is <math> \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{5}{1024} </math>. |
| − | If A wins 2 games, then B must win 2 or less games and the probability of this is <math> \frac{{ | + | If A wins 2 games, then B must win 2 or less games and the probability of this is <math> \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} </math>. |
| − | If A wins 3 games, then B must win 3 or less games and the probability of this is <math> \frac{{ | + | If A wins 3 games, then B must win 3 or less games and the probability of this is <math> \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} </math>. |
| − | If A wins 4 games, then B must win 4 or less games and the probability of this is <math> \frac{{ | + | If A wins 4 games, then B must win 4 or less games and the probability of this is <math> \frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024} </math>. |
| − | If A wins 5 games, then B must win 5 or less games and the probability of this is <math> \frac{{5 \choose 5}}{2^5} \frac{{ | + | If A wins 5 games, then B must win 5 or less games and the probability of this is <math> \frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024} </math>. |
Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so our answer is <math>319 + 512 = 831</math>. | Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so our answer is <math>319 + 512 = 831</math>. | ||
Revision as of 15:34, 12 March 2008
Problem
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a 
chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team 
beats team 
The probability that team finishes with more points than team 
is 
where 
and 
are relatively prime positive integers. Find 
Solution
You can break this into cases based on how many rounds A wins out of the remaining 5 games.
If A wins 0 games, then B must win 0 games and the probability of this is 
.
If A wins 1 games, then B must win 1 or less games and the probability of this is 
.
If A wins 2 games, then B must win 2 or less games and the probability of this is 
.
If A wins 3 games, then B must win 3 or less games and the probability of this is 
.
If A wins 4 games, then B must win 4 or less games and the probability of this is 
.
If A wins 5 games, then B must win 5 or less games and the probability of this is 
.
Summing these 6 cases, we get 
, which simplifies to 
, so our answer is 
.
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
