To solve this problem, we need to find the total number of possible solutions to the equation $ x + y + z = 10 $ where $ x, y, $ and $ z $ are nonnegative integers.

First, we employ the "stars and bars" method to determine the number of nonnegative integer solutions to this equation. This method states that the number of ways to partition $ n $ identical items (our sum) into $ k $ distinct groups (our variables) is given by:

$$ \binom{n + k - 1}{k - 1} $$

For $ x + y + z = 10 $, we have $ n = 10 $ and $ k = 3 $. Plugging into the formula, we get:

$$ \binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} $$

We calculate the binomial coefficient:

$$ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 $$

So, there are 66 possible solutions to the equation $ x + y + z = 10 $.

Next, we need to find the number of solutions for which $ z $ is even. Let $ z = 2k $ where $ k $ is a nonnegative integer. Then the equation becomes:

$$ x + y + 2k = 10 $$

Rearranging it, we get:

$$ x + y = 10 - 2k $$

Here, $ 10 - 2k $ must be nonnegative, so $ k $ can take values $ 0, 1, 2, 3, 4, 5 $ making total 6 possible values of k.

For each value of $ k $, $ x + y $ must equal the corresponding $ 10 - 2k $. The number of nonnegative integer solutions to $ x + y = m $ for any nonnegative integer $ m $ is given by:

$$ \binom{m + 1}{1} = m + 1 $$

We will sum the solutions for each valid $ k $:

When $ k = 0 $: $ x + y = 10 $, the number of solutions is $ 11 $.

When $ k = 1 $: $ x + y = 8 $, the number of solutions is $ 9 $.

When $ k = 2 $: $ x + y = 6 $, the number of solutions is $ 7 $.

When $ k = 3 $: $ x + y = 4 $, the number of solutions is $ 5 $.

When $ k = 4 $: $ x + y = 2 $, the number of solutions is $ 3 $.

When $ k = 5 $: $ x + y = 0 $, the number of solutions is $ 1 $.

Summing these, we get:

$$ 11 + 9 + 7 + 5 + 3 + 1 = 36 $$

Thus, there are 36 solutions where $ z $ is even out of a total of 66 solutions. Hence, the probability that $ z $ is even is given by:

$$ \frac{36}{66} = \frac{6}{11} $$

Thus, the correct answer is:

Option C: $$ \frac{6}{11} $$