2021 AIME I Problems/Problem 1

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 1 (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

Solution 2 (Casework but Bashier)

We have $5$ cases, depending on which race Zou lost. Let $W$ denote a won race, and $L$ denote a lost race for Zou. The possible cases are $WWWWL, WWWLW, WWLWW, WLWWW, LWWWW$ . The first case has probability $\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}$ . The second case has probability $\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}$ . The third has probability $\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}$ . The fourth has probability $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}$ . Lastly, the fifth has probability $\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}$ . Adding these up, the total probability is $\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}$ , so $m+n = \boxed{97}$ . ~rocketsri

This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few.

Solution 3 (Even more Casework)

Case 1: Zou loses the first race <\b>

In this case, Zou must win the rest of the races. Thus, our probability is $\frac{8}{243}$ .

Case 2: Zou loses the last race <\b>

There is only one possibility for this, so our probability is $\frac{16}{243}$ .

Case 3: Neither happens

There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}$ . Thus, the total probability is $\frac{8}{243} \cdot 3 = \frac{24}{243}$ .

Adding these up, we get $\frac{48}{243} = \frac{16}{81}$ .

Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY
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.

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2021 AIME I Problems/Problem 1

Contents

Problem

Solution 1 (Casework)

Solution 2 (Casework but Bashier)

Solution 3 (Even more Casework)

Video Solution by Punxsutawney Phil

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