2014 AMC 8 Problems/Problem 12

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

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 $3!\cdot \frac{1}{36}=\frac{1}{6}$ (B).

Alternate Solution:

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 $\dfrac{1}{3\cdot 2}=\boxed{\text{(B) }\dfrac{1}{6}}.$

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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2014 AMC 8 Problems/Problem 12

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