# 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 favorable responses are either they both show a green bean or they both show a red bean. 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 $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

## Solution 2

We can list out all the combinations and we get this: $GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2$. There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer $\boxed{\textbf{C}}$.

## Video Solution

https://youtu.be/NMpVIy8QxSY ~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 