Probability of choosing value of variable in equation - 4 tuple ($x_1 +x_2+x_3+x_4=10$) It may be the wording in this problem that is throwing me off but I can't seem to figure out the number of possible successful outcomes to calculate the probabiliy: 
Suppose a non-negative integer solution to the equation $w+x+y+z=10$ is chosen at random (each one being equally likely to be chosen). What is the probability that in this particular solution that $w$ is less than or equal to 2? 
Let $A = w \leq 2 $
To find P(A) I need: 
\begin{align}
P(A)=\frac{|E|}{|S|}
\end{align}
Where |E| = Successful outcomes and |S| = Size of sample space.
I start by finding the sample space of possible solutions, since this is a 4 tuple: ${\{w,x,y,z\}}$ --  order does not matter and repeats are allowed,  I would say the size of sample space is 
\begin{align}
|S| = C(10+4-1,4) =C(13,4)
\end{align}
So this gives me: 
\begin{align}
P(A)=\frac{|E|}{C(13,4)}
\end{align}
However, I can't seem to figure out $|E|$ as I don't know how to account for all cases... 
I am guessing since there are 4 variables: $\{w,x,y,z\}$ and we assume $w$ is aleady chosen from the following: $\{0,1,2\}$ (since $w\leq 2$) this leaves us with 3 variables left to determine. The number of outcomes for this would look like: 
$$
\begin{array}{c|lcr}
case & \text{Number of outcomes} \\
\hline
0+x+y+z= 10  & C(10+3-1,3) = C(12,3) \\
1+x+y+z= 10  & C(10+3-1,3) = C(12,3)\\
2+x+y+z= 10  & C(10+3-1,3) = C(12,3) 
\end{array}
$$
This feels wrong.. or maybe I am overthinking it. But would the solution be: 
\begin{align}
P(A)=\frac{3 \cdot C(12,3)}{C(13,4)}
\end{align}

Put w = 0
Find solution for $x+y+z = 10$ and that is ${12\choose2}=66$
Put w = 1
Find solution for $x+y+z = 9$  and that is ${11\choose2}=55$
Put w = 2
Find solution for $x+y+z = 8$  and that is ${10\choose2}=45$
Add the above cases to a total of $166$
For all non negative w,x,y,z 
Find the solution for $w+x+y+z = 10$ and that is ${13\choose3}
=286$
Thus the probability $=\frac{166}{286}$
