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Difference between revisions of "2017 AIME II Problems/Problem 2"
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Revision as of 11:50, 23 March 2017

Problem

The teams $T_1$, $T_2$, $T_3$, and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$, and $T_2$ plays $T_3$. The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$, the probability that $T_i$ wins is $\frac{i}{i+j}$, and the outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$, $T_3$ beats $T_2$, and $T_4$ beats $T_3$, and the second scenario is where $T_4$ beats $T_1$, $T_2$ beats $T_3$, and $T_4$ beats $T_3$. Consider the first scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{4+1}$, the probability $T_3$ beats $T_2$ is $\frac{3}{3+2}$, and the probability $T_4$ beats $T_3$ is $\frac{4}{4+3}$. Therefore the first scenario happens with probability $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}$. Consider the second scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{1+4}$, the probability $T_2$ beats $T_3$ is $\frac{2}{2+3}$, and the probability $T_4$ beats $T_2$ is $\frac{4}{4+2}$. Therefore the second scenario happens with probability $\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$. By summing these two probabilities, the probability that $T_4$ wins is $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$. Because this expression is equal to $\frac{256}{525}$, the answer is $256+525=\boxed{781}$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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