We know that the probability $P(k)$ of $k$ randomly chosen integers $(k \ge 2)$ from the set of natural number are coprime is
$$
P(k) = \frac{1}{\zeta(k)}.
$$
I am looking at a special case of this problem. Let $S_n$ be the set of all natural numbers which do not have a prime factor greater than $n$-th prime (i.e $S_n$ is the set of natural numbers that can be formed using only the first $n$ prime numbers). What is the probability $P(k, S_n)$ that $k$ randomly chosen integers $(k \ge 2)$ from the set $S_n$ are coprime?
I do not know the answer but I think it could be in a parametric form involving $n$ such that in the trivial case when $n\to \infty$, $P(k, S_{\infty}) = P(k) = 1/\zeta(k)$.
Edit: Explained the meaning of "all natural numbers."
Best Answer
First of all, note that there is no canonical notion of equidistribution on a countable set like the integers.
When asking for the probability that $k$ 'randomly chosen' integers are coprime, it is more-or-less intuitively clear what is meant by 'randomly chosen'. This is basically because the density of integers divisible by a given prime $p$ is (up to 'differences from rounding') the same in any interval $\{1, \dots, n\}$, namely $1/p$.
In your question this is not the case: for example in $S_4 \cap \{1, \dots, 10\}$, 5 of 10 numbers (= 50 percent) are even, while in $S_4 \cap \{1, \dots, 10^6\}$, 1070 of 1273 integers are even (which is about 84 percent) and in $S_4 \cap \{1, \dots, 10^{30}\}$, already 445064 of 462692 are even, which is about 96 percent.
So in order to make your question well-defined you need to be explicit about what you mean by 'randomly chosen'.