Percentage of all natural numbers divisible by a prime $p_n$ and no smaller prime The ratio of the count of all natural numbers divisible by $2$ to the count of all natural numbers is of course $\frac{1}{2}$ as there is no smaller prime than $2$.
The ratio of a count of all natural numbers divisible by $3$, but not by $2$ to the count of all natural numbers is $$\frac{1}{3}-\left(\frac{1}{2}\times\frac{1}{3}\right)=\frac{1}{6}$$
The ratio of a count of all natural numbers divisible by $5$, but not by $3$ or $2$ to the count of all natural numbers is $$\frac{1}{5}-\left(\frac{1}{2}\times\frac{1}{5}\right)-\left(\frac{1}{3}\times\frac{1}{5}\right)+\left(\frac{1}{2}\times\frac{1}{3}\times\frac{1}{5}\right)=\frac{1}{15}$$
The ratio of a count of all natural numbers divisible by $7$, but not by $5$, $3$ or $2$ to the count of all natural numbers is
$$\frac{1}{7}-\left(\frac{1}{2\ 7}+\frac{1}{3\ 7}+\frac{1}{5\ 7}-\frac{1}{2\ 3\ 7}-\frac{1}{2\ 5\ 7}-\frac{1}{3\ 5\ 7}+\frac{1}{2\ 3\ 5\ 7}\right)=\frac{4}{105}$$
This gets harder and harder to calculate with each larger prime. (I'm not completely sure this last one is correct)
Is there perhaps an easier way to think about this calculation to help find a general fraction formula for any prime, with the total sum of all fractions generated adding up to $1$?
 A: There's a much easier way to think about it. The natural density of natural numbers divisible by $p$ is $\frac{1}{p}$, so the density of numbers not divisible by $p$ is $1 - \frac{1}{p}$. Moreover the events of being divisible by different primes are "independent" in a suitable sense. So the density of natural numbers divisible by $p_n$ but not by $p_1, \dots p_{n-1}$ is
$$\left( \prod_{i=1}^{n-1} \left( 1 - \frac{1}{p_i} \right) \right) \frac{1}{p_n}.$$
Expanding this out produces the sum you get via inclusion-exclusion.
We need to be a little careful with this kind of "independence" reasoning because density is not actually a probability measure and so e.g. reasoning about infinitely many primes simultaneously this way is dodgy (although can, when done carefully, produce useful heuristics and conjectures), but you can prove rigorously that for finitely many primes everything works out. This is because the desired divisibility conditions are equivalent, via the Chinese remainder theorem, to a condition on the remainder $\bmod p_1 \dots p_n$, so the problem reduces to a counting problem in $\mathbb{Z}/(p_1 \dots p_n) \simeq \prod \mathbb{Z}/p_i$, which is finite.
