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Let $p$ be a prime number.
(a) What is the probability that the sum of $p$ non-negative integers is divisible by $p$?
(b) What is the probability that the product of $p$ non-negative integers is divisible by $p$?
Numbers are chosen randomly from $1$ to $p$ inclusive.

Ideas.
(a) Since there are $p$ residuals when dividing a number by $p$ and only when it is zero, the number is divisible, then the probability would be $\frac{1}{p}$. I have doubts in this case, since I consider that not all possible residuals are equiprobable for the sum.
(b) In this case, it is enough that a single number is divisible by $p$ so that the product is divisible as well. Then, the favorable cases would be $p^p$ (the ways to obtain the residuals for each factor) and the cases in which the product is not divisible would be when one with zero residual has not been taken, that is $(p-1) ^ p$ ways to take the numbers. So the probability would be $1-\left(\frac{p-1}{p}\right)^p.$

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    $\begingroup$ How are the $p$ integers chosen? (You can't have a uniform distribution over the integers--or the natural numbers.) $\endgroup$
    – paw88789
    Commented May 23, 2021 at 23:48
  • $\begingroup$ Numbers are chosen randomly, non-negative integers $\endgroup$
    – Rigo
    Commented May 24, 2021 at 0:02
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    $\begingroup$ @Rigo The point paw88789 is trying to make is that you can't actually choose a random nonnegative integer such that each number has an equal chance of being chosen. You could instead consider choosing from all nonnegative integers less than some large number $N$, compute the probability, and let $N \to \infty$; but this is not the same as "choosing uniformly from all nonnegative integers." $\endgroup$
    – angryavian
    Commented May 24, 2021 at 0:11
  • $\begingroup$ Thanks for your reply. It is true, the set from which they are selected must be delimited. Suppose that the integers are selected from the set from 1 to p, randomly. $\endgroup$
    – Rigo
    Commented May 24, 2021 at 0:22
  • $\begingroup$ With that clarification, both your answers are correct. $\endgroup$
    – Parcly Taxel
    Commented May 24, 2021 at 0:33

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