Every time these two wheels are spun, two numbers are selected by the pointers.  What is the probability that the sum of the two selected numbers is even?
$\text{(A)}\ \frac{1}{6} \qquad \text{(B)}\ \frac{3}{7} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{2}{3} \qquad \text{(E)}\ \frac{5}{7}$
For the [sum](https://artofproblemsolving.com/wiki/index.php/Sum) to be [even](https://artofproblemsolving.com/wiki/index.php/Even), the two selected numbers must have the same [parity](https://artofproblemsolving.com/wiki/index.php/Parity) (must be either odd and odd or even and even).  
The first spinner has $2$ odd numbers and $2$ even, so no matter what the second spinner is, there is a $1/2$ chance the first spinner lands on a number with the same parity, so the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) of an even sum is $1/2\rightarrow \boxed{\text{C}}$.