2
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.