Likelihood at least 2 out of $n$ numbers are visible to each other in $\mathbb{Z}^n$ Two points in $ \mathbb{Z}^n $  are said to be visible to each other, if they can be connected by a straight line, which doesn't intersect any points of $ \mathbb{Z}^n $ 
In Apostol's book "An introduction to Number Theory" an answer to the problem stated in the title was given for the special case $n = 2$
However I also wonder, if the problem could be generalized to $n > 2$, picking n points at random an checking, if there were at least $2$ points visible to each other. 
As always: Any constructive recommendation for literature, comment or answer is appreciated.    
 A: Partial answer: The probability for two points to be visible from each other in $n$ dimensions is $\frac1{\zeta(n)}$.
I'll interpret "picking two points at random" as follows. Since we're only interested in the relative position of the two points, I'll take the first one to be the origin. Then for any $R\in\Bbb N$ we count how many points in $[-R,R]^n$ are visible from the origin; call this number $N_R$. The quantity asked for is the limit $\lim_{R\to\infty}\frac{N_R}{(2R+1)^n}$. (There are $(2R+1)^n$ points we could have picked in $[-R,R]^n$.)
Note that $(x_1,\ldots,x_n)$ is visible from the origin iff $\gcd(x_1,\ldots,x_n)=1$. So $$\begin{align*}N_R&=(2R+1)^n-1-\#\{0\neq(x_1,\ldots,x_n)\in[-R,R]^n:\gcd(x_1,\ldots,x_n)>1\}\\
&=(2R+1)^n-1-\#\bigcup_{\substack{p\leq R\\p\text{ prime}}}\{(x_1,\ldots,x_n)\in[-R,R]^n:p\mid x_1,\ldots,x_n\}\\
&=(2R+1)^n-1+\sum_{k=2}^R\mu(k)\left(1+2\left\lfloor\frac Rk\right\rfloor\right)^n\qquad
\text{(by inclusion-exclusion)}\end{align*}$$
Note $$\left(1+2\left\lfloor\frac Rk\right\rfloor\right)^n=\left(\frac{2R}k+O(1)\right)^n=\left(\frac{2R}k\right)^n+O\left(\sum_{m=0}^{n-1}\binom nm\left(\frac{2R}k\right)^m\right)$$
and that $\sum_{m=0}^{n-1}\binom nm\left(\frac{2R}k\right)^m\leq n\left(\frac{2R}k\right)^{n-1}\max_m\binom nm=O_n\left(\left(\frac Rk\right)^{n-1}\right)$ so the probability for a point to be visible from the origin becomes
$$\begin{align*}\lim_{R\to\infty}\frac{N_R}{(2R+1)^n}&=1+\lim_{R\to\infty}\left(\frac{2R}{2R+1}\right)^n\sum_{k=2}^R\left(\frac{\mu(k)}{k^n}+O_n\left(\frac1{k^{n-1}R}\right)\right)\\
&=1+\lim_{R\to\infty}\sum_{k=2}^R\frac{\mu(k)}{k^n}+\lim_{R\to\infty}O_n\left(\int_1^R\frac{dx}{x^{n-1}R}\right)\\
&=1+\zeta(n)^{-1}-1+\lim_{R\to\infty}O_n\left(\int_1^R\frac{dx}{xR}\right)\\
&=\frac1{\zeta(n)}.\end{align*}$$
