<s>[INST] Find the probability of the event that $c=a+b$ 
In a set $\{0,1,2,3,\dots,n\}$, $3$ numbers $(a,b,c)$ are randomly chosen.
What is the probability of the event that $c=a+b$?

I thought, I should start with $c=0$ for which there is only $1$ way $c=0$, if both $a$ and $b$ are $0$.
For $c=1, 2$ ways: $a=1$ and $b=0$ or $a=0$ and $b=1$.
For $c=2, 3$ ways: $a=2$ and $b=0, a=0$ and $b=2$, $a=1$ and $b=1$,
and so on.
If we follow the pattern we can see that the number of outcomes for the terms provided is $c+1$.
I am stuck at the total number of outcomes. Since the set is from $0$ to $n$, logically the probability that $c=a+b$ should be $0$.
How can I continue this?
Thank you
 [/INST] You pointed out that given $c$, the number of ways to have $a+b=c$ is $c+1$. We now must figure out how many total ways there are to have $a+b=c$. This is a simple sum over the range of $c$:
$$\sum_{c=0}^n c+1$$
$$n+1 +\sum_{c=0}^n c$$
$$n+1 +\frac{n(n+1)}{2}$$
$$\frac{(n+2)(n+1)}{2}$$
To find the probability of this event occurring, we need only divide by the total number of possible outcomes, which is $(n+1)^3$:
$$\frac{(n+2)(n+1)}{2(n+1)^3}$$
$$\frac{(n+2)}{2(n+1)^2}$$
which is the answer.
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