Question about the probability of a number This is in reference to our video https://youtu.be/SpeDnV6pXsQ?list=PLhsb6tmzSpiwLds3DD62o1MvI2MS2bby7
which is Ram Murty's video on Analytic Number Theory.
On the 17th minute the professor says what is the probability that a random number is divisible by $p$?
and he himself answers that the probability is $\frac{1}{p}$?
However I dont understand how he found that?
Am I missing something ?Is the answer so much easy?
 A: It's a bit fuzzy (because just saying "if we pick a random integer" is fraught with trouble), but I would assume that he means something like this:
Pick a positive integer N, and let us draw an integer with uniform probability from the interval $[0, pN - 1]$. Then the numbers that are divisible by $p$ are $0, p, 2p, \ldots, p(N-1)$, and there are $N$ of them. So the fraction of values that are divisible by $p$ is $\frac{N}{pN} = \frac{1}{p}$, so the probability that our chosen number is divisible by $p$ is $\frac{1}{p}$. This doesn't depend on $N$, so as $N \rightarrow \infty$ the probability remains $\frac{1}{p}$.
(Note that you could just pick from the interval $[0, N - 1]$ and note that the probability fluctuates around $\frac{1}{p}$ with a decreasing error term, but this makes the math a little easier to do.)
Alternatively, you can just say that the density of the multiples of $p$ among the integers is $\frac{1}{p}$ which is proven the same way but which doesn't have the pitfalls of trying to work with probabilities on an infinite list.
A: Consider the numbers in the interval $[0,pn)$ for some large integer $n$.
Exactly $n$ of these numbers are divisible by $p$:
$$0,p,2p,3p,\dots,(n-1)p$$
So the proportion is
$$\frac{n}{np}=\frac1p.$$
