Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question

Three randomly chosen non negative integers $x, y  \text{  and  } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?

My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?

$x$ and $y$ need to have the same parity, so that we can predict that the probability will be close to $\frac12$.
For a given $z$, there are $11-z$ combinations of $x,y$, which are all of the same parity or of the opposite parity.
Hence 
$$p=\frac{11+9+7+5+3+1}{11+9+7+5+3+1+10+8+6+4+2}=\frac{36}{66}.$$

Using summation formulas, 
$$p=\frac{\left(\dfrac{12}2\right)^2}{\dfrac{11\cdot12}2}$$
