The probability that both show a green bean is $\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}$. The probability that both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$. 
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? (2013 AMC 8 Problem, Question \#14) 
Options:
a: $\dfrac{1}{4}$ 
b: $\dfrac{1}{3}$ 
c: $\dfrac{3}{8}$ 
d: $\dfrac{1}{2}$ 
e: $\dfrac{2}{3}$ 