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== | + | ==Solution (Casework)== |
| − | <math>\ | + | For the last five races, Zou wins four and loses one. There are five possible outcome sequences, and we will proceed by casework: |
| + | |||
| + | <b><u>Case (1): Zou does not lose the last race.</u></b> | ||
| + | |||
| + | There are four such outcome sequences. The probability of one such sequence is <math>\left(\frac23\right)^3\left(\frac13\right)^2.</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
==See also== | ==See also== | ||
Revision as of 17:21, 11 March 2021
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 (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 
Case (2): Zou loses the last race.
There is one such outcome sequence. The probability is 
Answer
The requested probability is ![\[4\left(\frac23\right)^3\left(\frac13\right)^2+\left(\frac23\right)^4\left(\frac13\right)^1=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\]](http://latex.artofproblemsolving.com/7/d/7/7d7d0609b19157f6819398fd8fc1e30eb63a32c2.png)
and the answer is 
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. 
