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Difference between revisions of "2017 AIME II Problems/Problem 2"

(Created page with "<math>\textbf{Problem 2}</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</...")
 
(→‎Solution: signed.)
 
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<math>\textbf{Problem 2}</math>
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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>.
  
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<math>\textbf{Problem 2 Solution}</math>
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==Solution 1==
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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>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>.
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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>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>.
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==Solution 2==
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Clearly <math>T_4</math> has to win its game with <math>T_1</math>, which has probability <math>\frac{4}{5}</math>. There are two cases, depending on who its opponent is.
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Case 1: <math>T_4</math> faces <math>T_2</math>. So <math>T_2</math> won its first game with probability <math>\frac{2}{5}</math>, and <math>T_4</math> wins the finals with probability <math>\frac{4}{6}=\frac{2}{3}</math>.
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Case 2: <math>T_4</math> faces <math>T_3</math>. So <math>T_3</math> won its first game with probability <math>\frac{3}{5}</math>, and <math>T_4</math> wins the finals with probability <math>\frac{4}{7}</math>.
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The total probability is therefore <math>\frac{4}{5} \times \left(\frac{4}{15} + \frac{12}{35}\right) = \frac{4}{5} \cdot \frac{64}{105} = \frac{256}{525} \implies 256+525 = \boxed{781}</math>.
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~First
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=See Also=
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{{AIME box|year=2017|n=II|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 14:25, 22 July 2025

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 1

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 $T_2$. 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 $256+525=\boxed{781}$.

Solution 2

Clearly $T_4$ has to win its game with $T_1$, which has probability $\frac{4}{5}$. There are two cases, depending on who its opponent is. Case 1: $T_4$ faces $T_2$. So $T_2$ won its first game with probability $\frac{2}{5}$, and $T_4$ wins the finals with probability $\frac{4}{6}=\frac{2}{3}$. Case 2: $T_4$ faces $T_3$. So $T_3$ won its first game with probability $\frac{3}{5}$, and $T_4$ wins the finals with probability $\frac{4}{7}$.

The total probability is therefore $\frac{4}{5} \times \left(\frac{4}{15} + \frac{12}{35}\right) = \frac{4}{5} \cdot \frac{64}{105} = \frac{256}{525} \implies 256+525 = \boxed{781}$. ~First

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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