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>W</math> denote a won race, and <math>L</math> denote a lost race for Zou. The possible cases are <math>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{97}</math>. ~rocketsri | ||
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| + | 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. | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
Revision as of 09:51, 12 March 2021
Contents
Problem
Zou and Chou are practicing their 100-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. What is 
?
Solution 1(Casework)
For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework:
Case (1): Zou does not lose the last race.
The probability that Zou loses a race is 
and the probability that he wins the following race is 
For each of the three other races, the probability that Zou wins is 
There are four such outcome sequences. The probability of one such sequence is 
Case (2): 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 such outcome sequence. The probability is 
Answer
The requested probability is ![\[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\]](http://latex.artofproblemsolving.com/b/7/9/b7966ab33b4b64d044c744aa5d34154f969ec195.png)
and 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.
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY
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. 
