2013 AMC 8 Problems/Problem 14

Problem

Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?

$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$

Solution

The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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