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?

Using computer power:
sage: allCases = len( [ S for S in cartesian_product( [ [0..10], [0..10], [0..10] ] ) if sum(S) == 10 ] )
sage: goodCases = len( [ S for S in cartesian_product( [ [0..10], [0..10], [0..10] ] ) if sum(S) == 10 and S[2] % 2 == 0 ] )
sage: allCases
66
sage: goodCases
36

