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Difference between revisions of "2014 AMC 8 Problems/Problem 12"

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==Problem==
 
==Problem==
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A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly? a
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A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly?  
  
 
<math> \textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{6}\qquad\textbf{(C) }\frac{1}{4}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2} </math>
 
<math> \textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{6}\qquad\textbf{(C) }\frac{1}{4}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2} </math>
  
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==Solution==
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==Solution 1==
 
Let's call the celebrities A, B, and C.  
 
Let's call the celebrities A, B, and C.  
 
There is a <math>\frac{1}{3}</math> chance that celebrity A's picture will be selected, and a <math>\frac{1}{3}</math> chance that his baby picture will be selected. That means there are two celebrities left. There is now a <math>\frac{1}{2}</math> chance that celebrity B's picture will be selected, and another <math>\frac{1}{2}</math> chance that his baby picture will be selected. This leaves a <math>\frac{1}{1}</math> chance for the last celebrity, so the total probability is <math>\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{36}</math>. However, the order of the celebrities doesn't matter, so the final probability will be <math> 3!\cdot \frac{1}{36}=\boxed{\text{(B) }\dfrac{1}{6}}.</math>
 
There is a <math>\frac{1}{3}</math> chance that celebrity A's picture will be selected, and a <math>\frac{1}{3}</math> chance that his baby picture will be selected. That means there are two celebrities left. There is now a <math>\frac{1}{2}</math> chance that celebrity B's picture will be selected, and another <math>\frac{1}{2}</math> chance that his baby picture will be selected. This leaves a <math>\frac{1}{1}</math> chance for the last celebrity, so the total probability is <math>\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{36}</math>. However, the order of the celebrities doesn't matter, so the final probability will be <math> 3!\cdot \frac{1}{36}=\boxed{\text{(B) }\dfrac{1}{6}}.</math>
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==Solution 3==
 
==Solution 3==
 
There are <math>3!</math> ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is <math>\boxed{\textbf{(B)}~\frac{1}{6}}</math>.
 
There are <math>3!</math> ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is <math>\boxed{\textbf{(B)}~\frac{1}{6}}</math>.
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/2SyV2oO_fOU
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/zoyYtCF4waw ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=11|num-a=13}}
 
{{AMC8 box|year=2014|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:12, 2 July 2023

Problem

A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly?

$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{6}\qquad\textbf{(C) }\frac{1}{4}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2}$

Solution 1

Let's call the celebrities A, B, and C. There is a $\frac{1}{3}$ chance that celebrity A's picture will be selected, and a $\frac{1}{3}$ chance that his baby picture will be selected. That means there are two celebrities left. There is now a $\frac{1}{2}$ chance that celebrity B's picture will be selected, and another $\frac{1}{2}$ chance that his baby picture will be selected. This leaves a $\frac{1}{1}$ chance for the last celebrity, so the total probability is $\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{36}$. However, the order of the celebrities doesn't matter, so the final probability will be $3!\cdot \frac{1}{36}=\boxed{\text{(B) }\dfrac{1}{6}}.$

Solution 2

There is a $\frac{1}{3}$ chance that the reader will choose the correct baby picture for the first person. Next, the second person gives a $\frac{1}{2}$ chance, and the last person leaves only 1 choice. Thus, the probability is $\dfrac{1}{3\cdot 2}=\boxed{\text{(B) }\dfrac{1}{6}}.$

Solution 3

There are $3!$ ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is $\boxed{\textbf{(B)}~\frac{1}{6}}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/2SyV2oO_fOU

~Education, the Study of Everything



Video Solution

https://youtu.be/zoyYtCF4waw ~savannahsolver

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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