<s>[INST] 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?
 [/INST] Presumably the intended problem is:


*
If $(x,y,z)$ is randomly chosen from the set of nonnegative integer triples $(x,y,z)$ satisfying $$x+y+z=10$$
what is the probability that $z$ is even?



Using that interpretation . . .

Suppose $x+y+z=10$, and $z$ is even.

Write $z=2c$.

Note that $x,y$ are either both even, or both odd.

If $x,y$ are both even, write $x=2a,\,y=2b$, and count the solutions to
$$2a+2b+2c=10$$
or equivalently,
$$a+b+c=5$$
If $x,y$ are both odd, write $x=2a+1,\,y=2b+1$, and count the solutions to
$$(2a+1)+(2b+1)+2c=10$$
or equivalently,
$$a+b+c=4$$

Then just sum  those counts, and divide by $66$.
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