I heard this question a few days ago, so reciting from memory:
If I were to randomly choose an arbitrarily large positive integer $n$, could I write a function that determines the likelihood of it being prime?
Intuitively, what would it mean to assign a probability to an integer of being prime?
Edit 1
I'm not sure how to incorporate the prime-counting function into this.
Edit 2
Alright, so the page on the prime number theorem says that:
Informally speaking, the prime number theorem states that if a random integer is selected in the range of zero to some large integer N, the probability that the selected integer is prime is about 1 / ln(N), where ln(N) is the natural logarithm of N.
Looking through the references, though, I can't find a more formal proof. So I will revise my question.
For a random integer selected in the range of $0$ to some large integer $N$, prove that the probability the selected integer is prime is $\frac{1}{\ln{N}}$.
Edit 3
If the selected integer $n$ was prime, it's necessary that it has no prime factors $p\leq\sqrt{n}$. If we could find the probability that $n$ is divisible by $p$ as some function of $p$, then could we write$$\prod_{p=2}^{\sqrt{n}}\left(1-f(p)\right)$$where $f(p)$ is the probability that $n$ is divisible by $p$?
Care should be taken in the sense we take the "density" of primes.
The prime number theorem states that $$ \pi(n)=\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\tag{1} $$ Thus, $$ \begin{align} \pi(n(1+\alpha))-\pi(n) &=\frac{n(1+\alpha)}{\log(n)+\log(1+\alpha)}-\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\\ &=\frac{n(1+\alpha)}{\log(n)}-\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\\ &=\frac{\alpha n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\tag{2} \end{align} $$ Therefore, $$ \frac{\pi(n(1+\alpha))-\pi(n)}{\alpha n} =\frac1{\log(n)}+O\left(\frac1{\alpha\log(n)^2}\right)\tag{3} $$ In the sense of $(3)$, the density of primes is $\dfrac1{\log(n)}$.