Difference between revisions of "2017 AIME II Problems/Problem 2"
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− | 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> | + | 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_2</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= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=1|num-a=3}} | {{AIME box|year=2017|n=II|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:57, 24 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 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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