AMC 8 1994 Problem 22

The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is

$\text{(A)}\ \dfrac{1}{6} \text{(B)}\ \dfrac{1}{4} \text{(C)}\ \dfrac{1}{3} \text{(D)}\ \dfrac{5}{12} \text{(E)}\ \dfrac{4}{9}$