Infinite Probability Space A number is chosen at random from the infinite set of numbers $\{ 1, 2, 3, ... \}$.
What is the probability that it is divisible by 3 or 5 or both?
I have this so far:
$P(A) = $ Div. by 3
$P(B) = $ Div. by 5
$P(A \cap B) = $ Div. by both 3 and 5
Thus:


$P(A \cup B) = P(A) + P(B) - P(A \cap B)$


But how do I go about using limits to solve the question?
Thanks
 A: Technically you can't choose an integer "uniformly at random" from the infinite set $\{1,2,3,\ldots\}$.  However you can choose a uniform random integer from the finite set $\{1,2,\ldots,n\}$ and then ask what is your desired probability as $n \to \infty$.  In that case, it should hopefully be clear that the probability that a number is divisible by $3$ approaches $1/3$, the probability that a number is divisible by $5$ approaches $1/5$, and the probability of being divisible by both approaches $1/15$. So by inclusion-exclusion, the overall probability is $1/3 + 1/5 - 1/15$.
A: A natural question that arises in connection to your question is how many numbers are even? I can think of infinitely many numbers that are even and I can think of infinitely many numbers that aren't even. So, that's not going to help much.
What if we look at proportions? Out of first 10 natural numbers, 5 are even. Out of first 20 natural numbers 10 are even. In fact the proportion is constant. Whenever we add 10 more numbers, 5 of them will be even. So 1/2 of all numbers are even. It makes sense to define the probability that a number is even as 1/2, or 50%.
