TITLE: Probability of selecting two numbers with a sum of squares divisible by 10
QUESTION [9 upvotes]: Two natural numbers $x$ and $y$ are chosen at random.  Find the probability that $x^2 + y^2$ is divisible by 10.

I could not understand how select two numbers from any natural number (infinite).

REPLY [9 votes]: You are partially right, and partially wrong.
It is impossible to define a uniform probability distribution on the natural numbers (i.e. one where each number has an equal probability of selection).
It is, however, possible to define non-uniform probability distributions on the sample space of all natural numbers. For example, roll a die until you roll a 6, and count how many rolls it took.
For the purpose of this question, you have to kind of take the position of "if there was a uniform distribution over $\mathbb{N}$, and you selected two numbers at random from it, what would be the probability of this property?", which does have a meaningful interpretation - take the sequence of distributions $U_n \sim \mbox{Uniform}(n)$, i.e. where the probability that $U_n = k$ for $k \in \{1, \ldots, n\}$ is $\frac{1}{n}$. Then take two draws $x_n$ and $y_n$ from $U_n$, and consider the probability that $x_n^2 + y_n^2$ is divisible by 10.
Then, consider what happens in the limit as $n \rightarrow \infty$. Does that probability converge to some fixed value? In fact, can you cheat and do the calculation as if the uniform distribution over the naturals existed in the first place? As in, if you do the calculation with $Prob(U_\infty = k) = \frac{1}{\delta}$ for an infinitesimal $\delta$, do all the $\delta$s drop out to give you something vaguely meaningful in the end? The answer is yes, and it's what the other answers to this question are leading you to.
(It would not be guaranteed that the probability would be the same for an arbitrary probability distribution over the natural numbers, which is why you have to assume the somewhat degenerate distribution I described exists.)