probability that two randomly chosen numbers are coprime Is this question well posed? See here for the solution Probability that two random numbers are coprime
I have also seen it in some contests. The question asks to compute $p=\lim p_n$ where $p_n$ is the probability that two random chosen integers less than $n$ are coprime. There is no way to associate a uniform distribution to integers; so I would hesitate to call this limit a probability. So is there any rigorous way to understand this limit as a probability of some event?
See also this post 
What's the mean of all real numbers?
where it is mentioned (and I agree) that the mean of reals (or integers) is undefined. But one could in the same way define a uniform distribution for reals or integers with absolute value less than $x>0$ and take the limit of the mean as $x$ goes to infinity. Then the mean of reals would be $0$. 
 A: Yes. Define a measure of a set $\Omega$ in the plane to be the limit of number of relatively prime points in $t \Omega$ divided by $t^2,$ as $t$ goes to infinity. It can be shown that for "nice" sets this is a measure, which is a multiple of the Lebesgue measure. The multiple is precisely the limit you talk about ($6/\pi^2$). You can look for a discussion along these lines in this paper if you want to know more.
A: Let us start with the following observation: 

One integer chosen amongst $'p'$ other integers has one chance to be divisible by $p$ 



*

*From this we infer that the probability that  an integer is divisible by $p$ is $\frac{1}{p}$.

*Therefore the probability that two different integers are both simultaneously divisible by a prime $p$ is $\frac{1}{p^2}$

*This means that the probability that two different integers are  not simultaneously divisible by a prime $p$ is $$1-\frac{1}{p^2}$$



  
*Conclusion: The probability that two different integers are  never simultaneously divisible by a prime (meaning that they are co-prime) 
  is then given by
  $$ \color{red}{\prod_{p, prime}\left(1-\frac{1}{p^2} \right) =
\left(\prod _{p, prime}\frac {1}{1-p^{-2}}\right)^{-1}=\frac {1}{\zeta (2)}=\frac {6}{\pi ^{2}} \approx  0,607927102 ≈ 61 \%}$$
  

Where should recall from the Basel problem we have the following Euler identity 
$$\frac{\pi^2}{6}=\sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2)=\prod _{p, prime}\frac {1}{1-p^{-2}} $$
By similar Token, The probability that $m$ numbers are co-prime is given by
$$ \color{red}{\prod_{p, prime}\left(1-\frac{1}{p^m} \right) =
\left(\prod _{p, prime}\frac {1}{1-p^{-m}}\right)^{-1}=\frac {1}{\zeta (m)}}$$ 
Here $\zeta$ is the Riemann zeta function. $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$
