Difference between revisions of "2021 AIME I Problems/Problem 1"
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== Solution 2 (Casework but Bashier) == | == Solution 2 (Casework but Bashier) == | ||
− | We have <math>5</math> cases, depending on which race Zou lost. Let <math>\text{W}</math> denote a won race, and <math>\text{L}</math> denote a lost race for Zou. The possible cases are <math>\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}</math>. The first case has probability <math>\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}</math>. The second case has probability <math>\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}</math>. The third has probability <math>\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}</math>. The fourth has probability <math>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}</math>. Lastly, the fifth has probability <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}</math>. Adding these up, the total probability is <math>\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}</math>, so <math>m+n = \boxed{ | + | We have <math>5</math> cases, depending on which race Zou lost. Let <math>\text{W}</math> denote a won race, and <math>\text{L}</math> denote a lost race for Zou. The possible cases are <math>\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}</math>. The first case has probability <math>\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}</math>. The second case has probability <math>\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}</math>. The third has probability <math>\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}</math>. The fourth has probability <math>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}</math>. Lastly, the fifth has probability <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}</math>. Adding these up, the total probability is <math>\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}</math>, so <math>m+n = \boxed{097}</math>. |
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+ | ~rocketsri | ||
This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few. | This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few. | ||
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==Solution 4 (Observations)== | ==Solution 4 (Observations)== | ||
− | Note that | + | Note that Zou wins one race. The probability that he wins the last race is <math>\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.</math> Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any <math>4</math> of the middle races for Zou to win. So the probability for this case is <math>4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.</math> Thus, the answer is <math>\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.</math> |
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+ | ~pinkpig | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
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https://youtu.be/F21t0PAzhLM | https://youtu.be/F21t0PAzhLM | ||
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+ | ==Video Solution by Power of Logic== | ||
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+ | https://youtu.be/WS6X1MQ37jg | ||
==See Also== | ==See Also== |
Latest revision as of 23:58, 22 August 2022
Contents
Problem
Zou and Chou are practicing their -meter sprints by running races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is if they won the previous race but only if they lost the previous race. The probability that Zou will win exactly of the races is , where and are relatively prime positive integers. Find
Solution 1 (Casework)
For the next five races, Zou wins four and loses one. Let and denote a win and a loss, respectively. There are five possible outcome sequences for Zou:
We proceed with casework:
Case (1): Sequences #1-4, in which Zou does not lose the last race.
The probability that Zou loses a race is and the probability that Zou wins the next race is For each of the three other races, the probability that Zou wins is
There are four sequences in this case. The probability of one such sequence is
Case (2): Sequence #5, in which Zou loses the last race.
The probability that Zou loses a race is For each of the four other races, the probability that Zou wins is
There is one sequence in this case. The probability is
Answer
The requested probability is from which the answer is
~MRENTHUSIASM
Solution 2 (Casework but Bashier)
We have cases, depending on which race Zou lost. Let denote a won race, and denote a lost race for Zou. The possible cases are . The first case has probability . The second case has probability . The third has probability . The fourth has probability . Lastly, the fifth has probability . Adding these up, the total probability is , so .
~rocketsri
This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few.
Solution 3 (Even More Casework)
Case 1: Zou loses the first race
In this case, Zou must win the rest of the races. Thus, our probability is .
Case 2: Zou loses the last race
There is only one possibility for this, so our probability is .
Case 3: Neither happens
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is . Thus, the total probability is .
Adding these up, we get , so .
~mathboy100
Solution 4 (Observations)
Note that Zou wins one race. The probability that he wins the last race is Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any of the middle races for Zou to win. So the probability for this case is Thus, the answer is
~pinkpig
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY
Video Solution
https://youtu.be/M3DsERqhiDk?t=15
Video Solution by Steven Chen (in Chinese)
Video Solution by Power of Logic
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.