Difference between revisions of "2006 AIME II Problems/Problem 10"
I like pie (talk | contribs) m (→Solution 2) |
(→Solution 1) |
||
Line 9: | Line 9: | ||
Of these three cases (<math>|A| > |B|, |A| < |B|, |A|=|B|</math>), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). | Of these three cases (<math>|A| > |B|, |A| < |B|, |A|=|B|</math>), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). | ||
− | There are <math>{ | + | There are <math>{5\choose k}</math> ways to <math>A</math> to have <math>k</math> victories, and <math>{5\choose k}</math> ways for <math>B</math> to have <math>k</math> victories. Summing for all values of <math>k</math>, |
− | <center><math>1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} { | + | <center><math>1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.</math></center> |
Thus <math>p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}</math>. The desired probability is the sum of the cases when <math>|A| \ge |B|</math>, so the answer is <math>\frac{126}{512} + \frac{193}{512} = \frac{319}{512}</math>, and <math>m+n = \boxed{831}</math>. | Thus <math>p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}</math>. The desired probability is the sum of the cases when <math>|A| \ge |B|</math>, so the answer is <math>\frac{126}{512} + \frac{193}{512} = \frac{319}{512}</math>, and <math>m+n = \boxed{831}</math>. | ||
Revision as of 11:16, 11 March 2010
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
Solution 1
The results of the five remaining games are independent of the first game, so by symmetry, the probability that scores higher than in these five games is equal to the probability that scores higher than . We let this probability be ; then the probability that and end with the same score in these give games is .
Of these three cases (), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ways to to have victories, and ways for to have victories. Summing for all values of ,
Thus . The desired probability is the sum of the cases when , so the answer is , and .
Solution 2
You can break this into cases based on how many rounds wins out of the remaining games.
- If wins 0 games, then must win 0 games and the probability of this is .
- If wins 1 games, then must win 1 or less games and the probability of this is .
- If wins 2 games, then must win 2 or less games and the probability of this is .
- If wins 3 games, then must win 3 or less games and the probability of this is .
- If wins 4 games, then must win 4 or less games and the probability of this is .
- If wins 5 games, then 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 |