AIME 2021 I · 第 1 题

Solution 1 (Casework 🪅)

For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:

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

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

Answer

The requested probability is

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

~MRENTHUSIASM

Solution 2 (Casework but Bashier)

We have $5$ cases, depending on which race Zou lost. Let $\text{W}$ denote a won race, and $\text{L}$ denote a lost race for Zou. The possible cases are $\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$. 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 $m+n = \boxed{097}$.

~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 second 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 $\frac{48}{243} = \frac{16}{81}$, so $16+81=\boxed{097}$.

~mathboy100

Solution 4 (Observations)

Note that Zou wins one race. The probability that he wins the last race is $\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.$ Thus, the answer is $\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.$

~pinkpig

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=H17E9n2nIyY

Video Solution

https://youtu.be/M3DsERqhiDk?t=15

Video Solution by Steven Chen (in Chinese)

https://youtu.be/F21t0PAzhLM

Video Solution by Power of Logic

https://youtu.be/WS6X1MQ37jg