As has been pointed out in the comments, there is no probability distribution on the positive integers that assigns equal weight to every integer.
But let $S\subseteq \mathbb{N}$ be a set of positive integers, and for every positive integer $n$, let $S_n$ be the set of all $k\in S$ such that $k\le n$. Let $|S_n|$ be the number of elements in $S_n$. Then $\lim_{n\to\infty} \frac{|S_n|}{n},$ if it exists, can be viewed as a measure of how "large" $S$ is. By that criterion, the answer for $S=\{10\}$ is $0$. The answer for $S$ the even numbers is $1/2$, while the answer for the primes is $0$.
However, $\lim_{n\to\infty}\frac{|S_n|}{n}$ need not exist. Moreover, even if we restrict attention to subsets of $\mathbb{N}$ for which the limit exists, this limit is not a probability distribution.
On the reals, there is no probability distribution that gives equal weight to all intervals of (say) length $1$. But let's restrict attention to a specific interval, say $[a,b]$, and use the uniform distribution on this interval. Then the probability that a randomly chosen point in $[a,b]$ is rational turns out to be $0$. Almost all real numbers are irrational, indeed almost all real numbers are transcendental.
For a proof that the rationals form a negligibly small subset of, say, $[0,1]$, let $\epsilon>0$. The rationals in $[0,1]$ form a countable set, so they can be listed as $r_1,r_2,r_3,\dots$. Put an interval of width $\frac{\epsilon}{2^1}$ about $r_1$, an interval of width $\frac{\epsilon}{2^2}$ about $r_2$, and so on. The sum of the lengths of these intervals is $\epsilon$. So the rationals are a subset of sets of arbitrarily small measure.