2021 AIME I Problems/Problem 1


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. Find $m+n.$

Solution 1 (Casework)

For the next five races, Zou wins four and loses one. There are five possible outcome sequences for him:

  1. $LWWWW$
  2. $WLWWW$
  3. $WWLWW$
  4. $WWWLW$
  5. $WWWWL$

We proceed with casework:

Case (1): Sequences #1-4, in which Zou does not lose the last race.

The probability that Zou loses a race is $\frac13,$ and the probability that he wins the next race is $\frac13.$ For each of the three other races, the probability that he wins is $\frac23.$

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

Case (2): Sequence #5, in which Zou loses the last race.

The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that he wins is $\frac23.$

There is one sequence in this case. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$


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


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

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

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 $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}$. 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


Video Solution


Video Solution by Steven Chen (in Chinese)


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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