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

m (→‎Solution (Casework): Signed my name.)
(→‎Solution (Casework): Made the explanations more detailed.)
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<b><u>Case (1): Zou does not lose the last race.</u></b>
 
<b><u>Case (1): Zou does not lose the last race.</u></b>
  
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There are four such outcome sequences. The probability of one such sequence is <math>\left(\frac23\right)^3\left(\frac13\right)^2.</math>
+
The probability that Zou loses a race is <math>\frac13,</math> and the probability that he wins the following race is <math>\frac13.</math> For each of the three other races, the probability that Zou wins is <math>\frac23.</math>
 +
 
 +
There are four such outcome sequences. The probability of one such sequence is <math>\left(\frac13\right)^2\left(\frac23\right)^3.</math>
  
 
<b><u>Case (2): Zou loses the last race.</u></b>
 
<b><u>Case (2): Zou loses the last race.</u></b>
  
−
There is one such outcome sequence. The probability is <math>\left(\frac23\right)^4\left(\frac13\right)^1.</math>
+
The probability that Zou loses a race is <math>\frac13.</math> For each of the four other races, the probability that Zou wins is <math>\frac23.</math>
 +
 
 +
There is one such outcome sequence. The probability is <math>\left(\frac13\right)^1\left(\frac23\right)^4.</math>
  
 
<b><u>Answer</u></b>
 
<b><u>Answer</u></b>
  
−
The requested probability is <cmath>4\left(\frac23\right)^3\left(\frac13\right)^2+\left(\frac23\right)^4\left(\frac13\right)^1=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},</cmath> and the answer is <math>16+81=\boxed{097}.</math>
+
The requested probability is <cmath>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},</cmath> and the answer is <math>16+81=\boxed{097}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 17:33, 11 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 (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

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

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