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?
 A: 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 $66$.
A: You are missing the fact that a single even value of $z$ represents several different soluitons, not one.
With $z$ equal to 10, there is only one possible solution (0+0+10).
With $z$ equal to 8, there are 3 possible solutions (0+2+8, 1+1+8, 2+0+8).
With $z$ equal to 6, there are 5 possible solutions.
With $z$ equal to 4, there are 7 possible solutions.
With $z$ equal to 2, there are 9 possible solutions.
With $z$ equal to 0, there are 11 possible solutions.
In total you have 36 solutions with $z$ being even and 66 solutions in total. So the probability is 36/66 or 6/11.
A: $x$ 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, 
$$p=\frac{\left(\dfrac{12}2\right)^2}{\dfrac{11\cdot12}2}$$
A: Since $x$, $y$, and $z$ are distinct, the number of possible solutions here is $11+10+9+...+1=66$, corresponding to $z=0$, $z=1$, etc.  (Your binomial coefficient is correct, but your reader will likely not be sure the reasoning for it, so if you are asking whether your statement is complete, you might want to add a justification for it.) The number of cases where $z$ is even is then $11+9+7+5+3+1=36$, so yes, $z$ is even in $6/11$ of the possible cases.  (Again your count is correct, but the reader may be unsure about the reasoning.)
On the other hand, "randomly chosen nonnegative integer" is ambiguous since there is no (proper) uniform distribution on the set of nonnegative integers. You might want to work out an alternate solution where the unconditional distribution is proper, something like $Prob(n)=\frac{1}{2^{n+1}}$. The answer then will still be $6/11$ since each case will have $1/2^{13}$ as its unconditional probability.  If you really want to have fun, you could use other improper distributions besides uniform. For example, you could try the Jeffreys prior for the Poisson.  I'm guessing that will give you something close to $6/11$ but not exactly the same.
A: 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

A: This is in fact a composition problem. We want to find all compositions of $x$ and $y$ such that $z$ takes values in $0,2,4,6,8,10$.
Then we count All 2-Compositions of $0,2,4,6,8,10$, that is how can we add 2 numbers to get these values. 
Let $C(n,k,a,b)$ be all k-compositions of n using the numbers from a to b.
We want:
$\sum_{i \in (0,2,4,6,8,10)} C(i,2,0,i)$ = 1+3+5+7+9+11=36.
We know $C(n,k,0,n)=$ $\sum_{i=2}^{10} [{i-1 \choose 1}+2]+1=66$ since we need 2 extra to account for 0 and 1 at the end to account for $x=0$ and $y=0$.
So we get $\frac {36}{66}$ as required.
