Difference between revisions of "2021 AIME I Problems/Problem 1"
MRENTHUSIASM (talk | contribs) (→Solution) |
MRENTHUSIASM (talk | contribs) m (→Solution (Casework): Signed my name.) |
||
Line 16: | Line 16: | ||
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> | ||
+ | |||
+ | ~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 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 and the answer is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.