It is well known that every coprime arithmetic progression (AP) contains an infinite number of prime numbers.
And also that the probability of $2$ random integers being coprime is $\dfrac{6}{\pi^2}$.
What is the probability of $2$ integers chosen at random from an admissible AP are coprime?
Two arbitrary integers $a$ and $b$ are coprime if and only if there is no prime $p$ such that $p|x$ and $p|y$. These conditions are independent, so the probability a given $p$ divides $a$ and $b$ is $p^{-2}$. These conditions are independent as $p$ varies, so the probability $b$ and $b$ are coprime is $$ \prod_p\left(1-p^{-2}\right)=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}. $$ With a little care this probabilistic argument can be made rigorous.
Let $k$ and $n$ be coprime integers. If we restrict $a$ and $b$ to be congruent to $k$ modulo $n$ (where $k$ and $n$ are coprime), then no prime dividing $n$ can divide $a$ or $b$. For $p\nmid n$, the probability $p$ divides $a$ and $b$ is unchanged. So, the probability $a$ and $b$ are coprime is $$ \prod_{p\nmid n}\left(1-p^{-2}\right)=\frac{6}{\pi^2}\prod_{p|n}\left(1-p^{-2}\right)^{-1}. $$