Probability that an integer is divisible by a prime $p$ Can any one explain why the probability that an integer is divisible by a prime $p$ (or any integer) is $1/p$?
 A: See http://en.wikipedia.org/wiki/Coprime_integers#Probabilities.
A: As I said in a comment, the notion of 'probability' over the set of all integers (or equivalently, the natural numbers) is fraught with some peril.  A better statement of the question is that the natural density of the numbers divisible by $p$ is $\frac{1}{p}$.  Natural density captures what people think of as probability; it simply represents the limit of the proportion of integers with the given property.  More specifically, the natural density of a set $A$ is defined as the limit $\lim_{n\rightarrow\infty}\frac{1}{n}\#\left\{i:i\leq n \wedge i\in A\right\}$.  For more details, see http://en.wikipedia.org/wiki/Natural_density.
In your particular case, the natural density result is easy to prove: the number of naturals $i\leq n$ that are divisible by $p$ (call this count $c$) satisfies $\frac{n}{p}-1\lt c\lt \frac{n}{p}+1$, so the density $d = \lim_{n\rightarrow\infty}\frac{c}{n}$ satisfies $\frac{1}{p}-\frac{1}{n}\lt d\lt \frac{1}{p}+\frac{1}{n}$ for all $n$; therefore we must have $d=\frac{1}{p}$.
A: Although Stephen Stadnicki's answer is much more rigorous, I only made sense of it with the following argument:
If $p$ is dividing a number $i$ in $1, \ldots, n$, then that number must be possible to write as $i = p \times j$, where $j \in \{1, \ldots, n/p\}$ ($n/p$ follows from the fact that $i \leq n$).
From the range of $j$, we can see that there are $n/p$ numbers in $1, \ldots, n$ divisible by $p$. Therefore, if we are taking a number uniformly from $1, \ldots, n$, there is a probability of $\frac{n/p}{n} = 1/p$ that it will be divisible by $p$.
