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

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(→‎Video Solution by Steven Chen (in Chinese))
 
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==Problem==
 
==Problem==
−
Zou and Chou are practicing their 100-meter sprints by running <math>6</math> races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is <math>\frac23</math> if they won the previous race but only <math>\frac13</math> if they lost the previous race. The probability that Zou will win exactly <math>5</math> of the <math>6</math> races is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
+
Zou and Chou are practicing their <math>100</math>-meter sprints by running <math>6</math> races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is <math>\frac23</math> if they won the previous race but only <math>\frac13</math> if they lost the previous race. The probability that Zou will win exactly <math>5</math> of the <math>6</math> races is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
 
==Solution 1 (Casework)==
 
==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:
+
For the next five races, Zou wins four and loses one. Let <math>W</math> and <math>L</math> denote a win and a loss, respectively. There are five possible outcome sequences for Zou:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>LWWWW</math></li><p>
 +
  <li><math>WLWWW</math></li><p>
 +
  <li><math>WWLWW</math></li><p>
 +
  <li><math>WWWLW</math></li><p>
 +
  <li><math>WWWWL</math></li><p>
 +
</ol>
 +
We proceed with casework:
  
−
<b><u>Case (1): Zou does not lose the last race.</u></b>
+
<b><u>Case (1): Sequences #1-4, in which Zou does not lose the last race.</u></b>
  
−
The probability that Zou loses a race is <math>\frac13,</math> and the probability that he wins the following race is <math>\frac13.</math> For each of the three other races, the probability that Zou wins is <math>\frac23.</math>
+
The probability that Zou loses a race is <math>\frac13,</math> and the probability that Zou wins the next race is <math>\frac13.</math> For each of the three other races, the probability that Zou wins is <math>\frac23.</math>
  
−
There are four such outcome sequences. The probability of one such sequence is <math>\left(\frac13\right)^2\left(\frac23\right)^3.</math>
+
There are four sequences in this case. The probability of one such sequence is <math>\left(\frac13\right)^2\left(\frac23\right)^3.</math>
  
−
<b><u>Case (2): Zou loses the last race.</u></b>
+
<b><u>Case (2): Sequence #5, in which Zou loses the last race.</u></b>
  
 
The probability that Zou loses a race is <math>\frac13.</math> For each of the four other races, the probability that Zou wins is <math>\frac23.</math>
 
The probability that Zou loses a race is <math>\frac13.</math> For each of the four other races, the probability that Zou wins is <math>\frac23.</math>
  
−
There is one such outcome sequence. The probability is <math>\left(\frac13\right)^1\left(\frac23\right)^4.</math>
+
There is one sequence in this case. The probability is <math>\left(\frac13\right)^1\left(\frac23\right)^4.</math>
  
 
<b><u>Answer</u></b>
 
<b><u>Answer</u></b>
  
−
The requested probability is <cmath>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},</cmath> and the answer is <math>16+81=\boxed{097}.</math>
+
The requested probability is <cmath>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},</cmath> from which the answer is <math>16+81=\boxed{097}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 
== 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{097}</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.
  
−
== Solution 3 (Even more Casework) ==
+
== Solution 3 (Even More Casework) ==
  
−
<b> Case 1: Zou loses the first race <\b>
+
<b> Case 1: Zou loses the first race </b>
  
 
In this case, Zou must win the rest of the races. Thus, our probability is <math>\frac{8}{243}</math>.
 
In this case, Zou must win the rest of the races. Thus, our probability is <math>\frac{8}{243}</math>.
  
−
<b> Case 2: Zou loses the last race <\b>
+
<b> Case 2: Zou loses the last race </b>
  
 
There is only one possibility for this, so our probability is <math>\frac{16}{243}</math>.
 
There is only one possibility for this, so our probability is <math>\frac{16}{243}</math>.
  
−
<b> Case 3: Neither happens <\b>
+
<b> Case 3: Neither happens </b>
  
−
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{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>, so <math>16+81=\boxed{097}</math>.
 +
 
 +
~mathboy100
 +
 
 +
==Solution 4 (Observations)==
 +
Note that Zou wins one race. The probability that he wins the last race is <math>\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.</math> 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 <math>4</math> of the middle races for Zou to win. So the probability for this case is <math>4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.</math> Thus, the answer is <math>\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.</math>
 +
 
 +
~pinkpig
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
 
https://youtube.com/watch?v=H17E9n2nIyY
 
https://youtube.com/watch?v=H17E9n2nIyY
  
−
==See also==
+
==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
 +
 
 +
==See Also==
 
{{AIME box|year=2021|n=I|before=First problem|num-a=2}}
 
{{AIME box|year=2021|n=I|before=First problem|num-a=2}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:58, 22 August 2022

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. 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 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 $\frac{48}{243} = \frac{16}{81}$, 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

See Also

2021 AIME I (Problems • Answer Key • Resources)
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First problem 		Followed by
Problem 2
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