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

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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(\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>
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~MRENTHUSIASM
  
 
==See also==
 
==See also==

Revision as of 16:23, 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.

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

Case (2): Zou loses the last race.

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

Answer

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

~MRENTHUSIASM

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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