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

  • 3
    $\begingroup$ 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?) $\endgroup$ – Did Sep 11 '16 at 19:42
  • $\begingroup$ Sorry, I meant total number of outcomes with sample space. No, I can not compute P(a+b<=n). $\endgroup$ – zeeks Sep 11 '16 at 20:00
  • 1
    $\begingroup$ " 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}? $\endgroup$ – 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 $c$:

$$\sum_{c=0}^n c+1$$

$$n+1 +\sum_{c=0}^n c$$

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


To find the probability of this event occurring, we need only divide by the total number of possible outcomes, which is $(n+1)^3$:



which is the answer.


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