# 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

• If a+b=k with k<=n, the probability that c=k is 1/(n+1) while if a+b=k with k>n, the probability that c=k is 0, hence the global probability that c=a+b is P(a+b<=n)/(n+1). Can you compute P(a+b<=n)? ("I am stuck at the sample space." Sorry but why should you care about the sample space?) – Did Sep 11 '16 at 19:42
• Sorry, I meant total number of outcomes with sample space. No, I can not compute P(a+b<=n). – zeeks Sep 11 '16 at 20:00
• " I can not compute P(a+b<=n)" Why is that? Can't you enumerate the relevant couples in {0,1,2,3,…,n}x{0,1,2,3,…,n}? – Did Sep 11 '16 at 20:01

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 $(n+1)^3$:

$$\frac{(n+2)(n+1)}{2(n+1)^3}$$

$$\frac{(n+2)}{2(n+1)^2}$$