For the sum of the two numbers removed to be even, they must be of the same [parity](https://artofproblemsolving.com/wiki/index.php/Parity).  There are five even values and five [odd](https://artofproblemsolving.com/wiki/index.php/Odd) values. 
No matter what Jack chooses, the number of numbers with the same parity is four.  There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is $\frac49$
$\boxed{\text{A}}$
Ten balls numbered $1$ to $10$ are in a jar.  Jack reaches into the jar and randomly removes one of the balls.  Then Jill reaches into the jar and randomly removes a different ball.  The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the [sum](https://artofproblemsolving.com/wiki/index.php/Sum) of the two numbers on the balls removed is [even](https://artofproblemsolving.com/wiki/index.php/Even) is 
$\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{10}{19} \qquad \text{(E)}\ \frac{5}{9}$