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Difference between revisions of "2021 AIME I Problems/Problem 1"

m (→‎Solution 1 (Casework))
m (Minor change to all 3 sols)
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For the next five races, Zou wins four and loses one. There are five possible outcome sequences for Zou:
 
For the next five races, Zou wins four and loses one. There are five possible outcome sequences for Zou:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
−
   <li><math>LWWWW</math></li><p>
+
   <li><math>\text{LWWWW}</math></li><p>
−
   <li><math>WLWWW</math></li><p>
+
   <li><math>\text{WLWWW}</math></li><p>
−
   <li><math>WWLWW</math></li><p>
+
   <li><math>\text{WWLWW}</math></li><p>
−
   <li><math>WWWLW</math></li><p>
+
   <li><math>\text{WWWLW}</math></li><p>
−
   <li><math>WWWWL</math></li><p>
+
   <li><math>\text{WWWWL}</math></li><p>
 
</ol>
 
</ol>
 
We proceed with casework:
 
We proceed with casework:
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== Solution 2 (Casework but Bashier) ==
 
== Solution 2 (Casework but Bashier) ==
−
We have <math>5</math> cases, depending on which race Zou lost. Let <math>W</math> denote a won race, and <math>L</math> denote a lost race for Zou. The possible cases are <math>WWWWL, WWWLW, WWLWW, WLWWW, LWWWW</math>. The first case has probability <math>\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}</math>. The second case has probability <math>\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}</math>. The third has probability <math>\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}</math>. The fourth has probability <math>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}</math>. Lastly, the fifth has probability <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}</math>. Adding these up, the total probability is <math>\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}</math>, so <math>m+n = \boxed{97}</math>. ~rocketsri
+
We have <math>5</math> cases, depending on which race Zou lost. Let <math>\text{W}</math> denote a won race, and <math>\text{L}</math> denote a lost race for Zou. The possible cases are <math>\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}</math>. The first case has probability <math>\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}</math>. The second case has probability <math>\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}</math>. The third has probability <math>\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}</math>. The fourth has probability <math>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}</math>. Lastly, the fifth has probability <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}</math>. Adding these up, the total probability is <math>\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}</math>, so <math>m+n = \boxed{97}</math>. ~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.
 
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.
Line 50: Line 50:
 
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}</math>. Thus, the total probability is <math>\frac{8}{243} \cdot 3 = \frac{24}{243}</math>.
 
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is <math>\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}</math>. Thus, the total probability is <math>\frac{8}{243} \cdot 3 = \frac{24}{243}</math>.
  
−
Adding these up, we get <math>\frac{48}{243} = \frac{16}{81}</math>.
+
Adding these up, we get <math>\frac{48}{243} = \frac{16}{81}</math>. <math>16+81=\boxed{097}</math>
  
 
~mathboy100
 
~mathboy100

Revision as of 20:00, 9 September 2021

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. Find $m+n.$

Solution 1 (Casework)

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

  1. $\text{LWWWW}$

  2. $\text{WLWWW}$

  3. $\text{WWLWW}$

  4. $\text{WWWLW}$

  5. $\text{WWWWL}$

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 $m+n = \boxed{97}$. ~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 $\frac{48}{243} = \frac{16}{81}$. $16+81=\boxed{097}$

~mathboy100

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

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

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