1. Total number of nonnegative integer solutions

We need the number of triples $(x,y,z)$ of nonnegative integers satisfying $x+y+z=10.$ Using stars and bars, the number of solutions is $\binom{10+3-1}{3-1}=\binom{12}{2}=66.$

2. Count solutions for which $z$ is even

Since $z$ is a nonnegative even integer and $x+y+z=10$, possible values of $z$ are $z=0,2,4,6,8,10.$

For each fixed $z$, the number of nonnegative integer solutions of $x+y=10-z$ is $10-z+1=11-z.$

Now count:

• If $z=0$, solutions: $11$
• If $z=2$, solutions: $9$
• If $z=4$, solutions: $7$
• If $z=6$, solutions: $5$
• If $z=8$, solutions: $3$
• If $z=10$, solutions: $1$

Total favorable solutions: $11+9+7+5+3+1=36.$

3. Compute the probability

Therefore, $P(z\text{ is even})=\frac{36}{66}=\frac{6}{11}.$

4. Match with options

$\frac{6}{11}$ corresponds to Option C.

5. Comparison with stored answer

Stored correct answer is C, which matches our derived result.