If you consider any multiple of $pq$ consecutive integers, say $\mathcal{S}=\{1,2,3,\dots,kpq\}$, exactly $\frac1{pq}$ of them will be divisible by both $p$ and $q$; exactly $\frac1p$ will be divisible by $p$, and exactly $\frac1q$ will be divisible by $q$. Thus, exactly $\frac1{p^2q^2}$ of the pairs in $\mathcal{S}\times\mathcal{S}$ will be divisible by both $p$ and $q$, exactly $\frac1{p^2}$ will be divisible by $p^2$, and exactly $\frac1{q^2}$ will be divisible by $q^2$.
This can be extended for any set of primes; just take a set of $pqr\dots s$ consecutive integers, $\mathcal{S}=\{1,2,3,\dots,kpqr\dots s\}$, exactly $\frac1{p^2q^2r^2\dots s^2}$ of the pairs in $\mathcal{S}\times\mathcal{S}$ will be divisible by all $p,q,r,\dots s$. Thus, we don't need to appeal to probability, and what would be considered independence in a probabilistic setting, follows from the Chinese Remainder Theorem.
Chinese Remainder Theorem
If you want to know how many integers are relatively prime to $pqr\dots s$ in the set $\mathcal{S}=\{1,2,3,\dots,kpqr\dots s\}$, the Chinese Remainder Theorem says that there is one solution to
$$
\begin{align}
x&\equiv a\pmod{p}\\
x&\equiv b\pmod{q}\\
x&\equiv c\pmod{r}\\
&\vdots\\
x&\equiv d\pmod{s}
\end{align}
$$
mod $pqr\dots s$ for each choice of $a,b,c,\dots,d$. Thus, to be relatively prime to $pqr\dots s$, we can choose $a,b,c,\dots,d$ to be any non-zero residue modulo the corresponding prime. That is, there are $(p-1)(q-1)(s-1)\dots(s-1)$ numbers mod $pqr\dots s$ that are relatively prime to $pqr\dots s$. In other words, $\frac{(p-1)(q-1)(r-1)\dots(s-1)}{pqr\dots s}$ of the numbers in $\mathcal{S}$ are relatively prime to $pqr\dots s$.
Limits
Define
$$
\mathcal{S}_n=\left\{j\in\mathbb{Z}:1\le j\le n\right\}
$$
Note that the number of pairs in $\mathcal{S}_n\times\mathcal{S}_n$ sharing any prime $p$ is at most $\frac{n^2}{p^2}$.
Choose an $\epsilon\gt0$ and then pick an $N$ so that
$$
\sum_{p\ge N}\frac1{p^2}\le\epsilon
$$
and choose a $K$ so that $K\epsilon\ge1$. Set $P=\prod\limits_{p\lt N}p$.
We know that exactly $\prod\limits_{p\lt N}\left(1-\frac1{p^2}\right)$ of the pairs in $\mathcal{S}_{kP}\times\mathcal{S}_{kP}$ share no prime factor less than $N$.
Consider $\mathcal{S}_n$ where $kP\le n\le(k+1)P$ and $k\ge K$. Since the ratio of the sizes of these sets is no more than $\frac{k+1}{k}$, the fraction of pairs can only change by a ratio of $\left(\frac{k+1}{k}\right)^2\le(1+\epsilon)^2$.
Furthermore, the fraction of pairs of $\mathcal{S}_n\times\mathcal{S}_n$ sharing any prime greater than $N$ is at most
$$
\frac1{n^2}\sum_{p\ge N}\frac{n^2}{p^2}\le\epsilon
$$
Therefore, because the variation in the fraction of pairs of $\mathcal{S}_n\times\mathcal{S}_n$ varies little and the fraction of pairs sharing primes unaccounted for is small, we get that the fraction of pairs of $\mathcal{S}_n\times\mathcal{S}_n$ that share no prime factors tends to
$$
\prod_{p\in\mathbb{P}}\left(1-\frac1{p^2}\right)=\left(\sum_{k=1}^\infty\frac1{k^2}\right)^{-1}=\frac6{\pi^2}
$$
