2021 AIME I Problem 1

Below is the professionally curated solution for Problem 1 of the 2021 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AIME I solutions, or check the answer key.

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Concepts:conditional probabilitycasework

Difficulty rating: 2050

1.

Zou and Chou are practicing their 100100-meter sprints by running 66 races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is 23\frac{2}{3} if they won the previous race but only 13\frac{1}{3} if they lost the previous race. The probability that Zou will win exactly 55 of the 66 races is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

Zou wins race 1,1, so winning exactly 55 of the 66 races means he loses exactly one of races 22 through 6.6. Each race after the first repeats the previous outcome with probability 23\frac{2}{3} and switches with probability 13.\frac{1}{3}.

If the loss is race 6,6, the five transitions are four repeats followed by one switch: (23)413=16243.\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{243}. If the loss is race ii for some 2i5,2 \le i \le 5, there is a switch into the loss and a switch back to winning, plus three repeats: (23)3(13)2=8243\left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = \frac{8}{243} for each of the 44 positions, contributing 32243.\frac{32}{243}.

The total is 16243+32243=48243=1681,\frac{16}{243} + \frac{32}{243} = \frac{48}{243} = \frac{16}{81}, so m+n=16+81=97.m + n = 16 + 81 = 97.

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