A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ uniformly and independently, the above probability tends to some number $p(n)$ as $N \rightarrow \infty$). 


In the above sense, what is the probability $p(n,k)$ that among n integers $x\_1, x\_2, \dots, x\_n$ chosen at random we have $\gcd (x\_{i\_1}, x\_{i\_2}, \dots, x\_{i\_k})=1$ for all possible combinations of the $x\_{i\_j}$ (with $s \neq t$ if for $u \neq v$, $x\_{i\_u}=x\_s$ and $x\_{i\_v}=x\_t$) ?



Probability of all combinations of k numbers among n being coprime
For any prime $q$, define
$$
\ell(q,n,k) = q^{-n} \sum\_{j=0}^{n-k} \binom nj (q-1)^j,
$$
so that $\ell(q,n,k)$ is the probability, if a biased coin that comes up heads only $1/q$ of the time is tossed $n$ times, that at least $k$ heads are obtained. Equivalently, $\ell(q,n,k)$ is the probability, if $n$ numbers are chosen uniformly and independently from the set $\{0,1,\dots,q-1\}$, that at least $k$ of the numbers equal $0$.


Then the same argument giving the $1/\zeta(n)$ result shows that the probability that the greatest common divisor of every $k$-subset of $n$ "randomly chosen" integers is $1$ equals
$$
\prod\_q \big( 1 - \ell(q,n,k) \big),
$$
where the product is over all primes $q$. This product is convergent as long as $k\ge2$; it diverges to $0$ if $k=1$ (appropriately).


For example, if three integers are chosen at random, the probability that they are pairwise coprime is
$$
\prod\_q \big( 1 - \ell(q,3,2) \big) = \prod\_q \bigg( 1-\frac1q \bigg)^2 \bigg( 1+\frac2q \bigg) \approx 0.286747.
$$
(A million trials with random integers between $1$ and $10^{60}$ yielded $286912$ pairwise coprime triples, so this limiting probability seems accurate.) I don't believe this product has a closed form in terms of the zeta function.


