Difference between revisions of "2017 AIME II Problems/Problem 2"
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− | + | ==Problem== | |
The teams <math>T_1</math>, <math>T_2</math>, <math>T_3</math>, and <math>T_4</math> are in the playoffs. In the semifinal matches, <math>T_1</math> plays <math>T_4</math>, and <math>T_2</math> plays <math>T_3</math>. The winners of those two matches will play each other in the final match to determine the champion. When <math>T_i</math> plays <math>T_j</math>, the probability that <math>T_i</math> wins is <math>\frac{i}{i+j}</math>, and the outcomes of all the matches are independent. The probability that <math>T_4</math> will be the champion is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | The teams <math>T_1</math>, <math>T_2</math>, <math>T_3</math>, and <math>T_4</math> are in the playoffs. In the semifinal matches, <math>T_1</math> plays <math>T_4</math>, and <math>T_2</math> plays <math>T_3</math>. The winners of those two matches will play each other in the final match to determine the champion. When <math>T_i</math> plays <math>T_j</math>, the probability that <math>T_i</math> wins is <math>\frac{i}{i+j}</math>, and the outcomes of all the matches are independent. The probability that <math>T_4</math> will be the champion is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
− | + | ==Solution== | |
There are two scenarios in which <math>T_4</math> wins. The first scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_3</math> beats <math>T_2</math>, and <math>T_4</math> beats <math>T_3</math>, and the second scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_2</math> beats <math>T_3</math>, and <math>T_4</math> beats <math>T_3</math>. Consider the first scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{4+1}</math>, the probability <math>T_3</math> beats <math>T_2</math> is <math>\frac{3}{3+2}</math>, and the probability <math>T_4</math> beats <math>T_3</math> is <math>\frac{4}{4+3}</math>. Therefore the first scenario happens with probability <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}</math>. Consider the second scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{1+4}</math>, the probability <math>T_2</math> beats <math>T_3</math> is <math>\frac{2}{2+3}</math>, and the probability <math>T_4</math> beats <math>T_2</math> is <math>\frac{4}{4+2}</math>. Therefore the second scenario happens with probability <math>\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. By summing these two probabilities, the probability that <math>T_4</math> wins is <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. Because this expression is equal to <math>\frac{256}{525}</math>, the answer is <math>256+525=\boxed{781}</math>. | There are two scenarios in which <math>T_4</math> wins. The first scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_3</math> beats <math>T_2</math>, and <math>T_4</math> beats <math>T_3</math>, and the second scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_2</math> beats <math>T_3</math>, and <math>T_4</math> beats <math>T_3</math>. Consider the first scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{4+1}</math>, the probability <math>T_3</math> beats <math>T_2</math> is <math>\frac{3}{3+2}</math>, and the probability <math>T_4</math> beats <math>T_3</math> is <math>\frac{4}{4+3}</math>. Therefore the first scenario happens with probability <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}</math>. Consider the second scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{1+4}</math>, the probability <math>T_2</math> beats <math>T_3</math> is <math>\frac{2}{2+3}</math>, and the probability <math>T_4</math> beats <math>T_2</math> is <math>\frac{4}{4+2}</math>. Therefore the second scenario happens with probability <math>\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. By summing these two probabilities, the probability that <math>T_4</math> wins is <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. Because this expression is equal to <math>\frac{256}{525}</math>, the answer is <math>256+525=\boxed{781}</math>. | ||
+ | |||
+ | =See Also= | ||
+ | {{AIME box|year=2017|n=II|num-b=0|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 12:50, 23 March 2017
Problem
The teams ,
,
, and
are in the playoffs. In the semifinal matches,
plays
, and
plays
. The winners of those two matches will play each other in the final match to determine the champion. When
plays
, the probability that
wins is
, and the outcomes of all the matches are independent. The probability that
will be the champion is
, where
and
are relatively prime positive integers. Find
.
Solution
There are two scenarios in which wins. The first scenario is where
beats
,
beats
, and
beats
, and the second scenario is where
beats
,
beats
, and
beats
. Consider the first scenario. The probability
beats
is
, the probability
beats
is
, and the probability
beats
is
. Therefore the first scenario happens with probability
. Consider the second scenario. The probability
beats
is
, the probability
beats
is
, and the probability
beats
is
. Therefore the second scenario happens with probability
. By summing these two probabilities, the probability that
wins is
. Because this expression is equal to
, the answer is
.
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 0 |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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