The probability of both show a green bean is $\dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{1}{8}$. The probability of both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore, the probability is $\dfrac{1}{4}+\dfrac{1}{8}=\boxed {\left (\text{C}\right )\dfrac{3}{8}}$. 
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 \# $14$) 

  - $\dfrac{1}{4}$ 
  - $\dfrac{1}{3}$ 
  - $\dfrac{3}{8}$ 
  - $\dfrac{1}{2}$ 
  - $\dfrac{2}{3}$ 