Difference between revisions of "2021 AIME I Problems/Problem 1"

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Zou and Chou are practicing their 100-meter sprints by running <math>6</math> races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is <math>\frac23</math> if they won the previous race but only <math>\frac13</math> if they lost the previous race. The probability that Zou will win exactly <math>5</math> of the <math>6</math> races is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
 
Zou and Chou are practicing their 100-meter sprints by running <math>6</math> races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is <math>\frac23</math> if they won the previous race but only <math>\frac13</math> if they lost the previous race. The probability that Zou will win exactly <math>5</math> of the <math>6</math> races is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
  
==Solution (Casework)==
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==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:
 
For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework:
  

Revision as of 10:43, 12 March 2021

Problem

Zou and Chou are practicing their 100-meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

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 $\frac13,$ and the probability that he wins the following race is $\frac13.$ For each of the three other races, the probability that Zou wins is $\frac23.$

There are four such outcome sequences. The probability of one such sequence is $\left(\frac13\right)^2\left(\frac23\right)^3.$

Case (2): Zou loses the last race.

The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that Zou wins is $\frac23.$

There is one such outcome sequence. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$

Answer

The requested probability is \[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\] and the answer is $16+81=\boxed{097}.$

~MRENTHUSIASM

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=H17E9n2nIyY

See also

2021 AIME I (ProblemsAnswer KeyResources)
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

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