Two numbers are relatively prime if they have no common divisors except 1. It also means that the numbers have no prime factor in common.

The probability that a number is divisible by $2$ is $1/2$.

The probability that two numbers are both divisible by $2$ is $(\frac{1}{2}) .\frac{1}{2} = \frac{1}{2^2}$.
And thus, probability that neither of the two numbers are divisible by $2$ is $(1 - \frac{1}{2^2})$

Similarly, probability that neither of the two numbers are divisible by $3$ is $(1-\frac{1}{3^2})$

Thus the probability that neither of the two numbers are divisible by any of the prime numbers is
$$
P(Relatively Prime) = (1-\frac{1}{2^2}).(1-\frac{1}{3^2}).(1-\frac{1}{5^2}).(1-\frac{1}{7^2})...
$$
**However**, this is the minimum probability limit for any two numbers being relatively prime.

"Wait, but that's an infinite product. What am I gonna do with that big product?"

We'll see.

Let's intruduce an infinite sum, $$ S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + ... $$ Multiplying both sides by $\frac{1}{2^2}$,

$$ \frac{1}{2^2}.S = \frac{1}{2^2}.\frac{1}{1^2} + \frac{1}{2^2}.\frac{1}{2^2} + \frac{1}{2^2}.\frac{1}{3^2} + \frac{1}{2^2}.\frac{1}{4^2} + \frac{1}{2^2}.\frac{1}{5^2} + ... $$ $$ \frac{1}{2^2}.S = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} +\frac{1}{8^2} +\frac{1}{10^2} + ... $$

Subtracting the last term from $S$, $$(1 - \frac{1}{2^2}).S = 1 + (\frac{1}{2^2} - \frac{1}{2^2}) + \frac{1}{3^2} + (\frac{1}{4^2} - \frac{1}{4^2}) + \frac{1}{5^2} + ...$$

$$S_1 = (1 - \frac{1}{2^2}).S = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} ...$$

All the multiples of 2 are gone.

Now, multiply $S_1$ by $\frac{1}{3^2}$. $$ \frac{1}{3^2}.S_1 = \frac{1}{3^2}.1 + \frac{1}{3^2}.\frac{1}{3^2} + \frac{1}{3^2}.\frac{1}{5^2} + \frac{1}{3^2}.\frac{1}{7^2} ... $$

Subtracting the last term from $S_1$. $$ (1 - \frac{1}{3^2}).S_1 = 1 + \frac{1}{5^2} + \frac{1}{7^2} + (\frac{1}{9^2} - \frac{1}{9^2}) + ... $$ $$ (1 - \frac{1}{3^2}).S_1 = 1 + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{11^2} + ... + \frac{1}{23^2} + \frac{1}{25^2} + ... $$ $$ (1 - \frac{1}{3^2}).(1 - \frac{1}{2^2}).S = 1 + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{11^2} + ... + \frac{1}{23^2} + \frac{1}{25^2} + ... $$

All the remaining multiples of 3 are gone.

If we continue this with all the prime numbers, all the terms on the right are gone except 1. $$ S.(1 - \frac{1}{2^2}).(1 - \frac{1}{3^2})(1 - \frac{1}{5^2})(1 - \frac{1}{7^2})(1 - \frac{1}{11^2})... = 1 $$ So, $$ S = \frac{1}{(1 - \frac{1}{2^2}).(1 - \frac{1}{3^2})(1 - \frac{1}{5^2})(1 - \frac{1}{7^2})(1 - \frac{1}{11^2})...} $$ $$ (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{5^2})(1 - \frac{1}{7^2})(1 - \frac{1}{11^2})... = \frac{1}{S} $$ At the beginning, we defined $$ S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + ... $$ This is a very famous sum defined as, $$ \zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + ... $$ The very beautiful symbol $\zeta$ is called Reimann Zeta Function named after the mathematician Bernhard Riemann. It is defined for all complex numbers as, $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} + ... $$ where $s$ is a complex number.

It might be surprising, but it is true that $$ \zeta(2) = \frac{\pi^2}{6} $$ The proof to this fact is here.

So, going back to our original question, $$ P(Relatively Prime) = (1-\frac{1}{2^2}).(1-\frac{1}{3^2}).(1-\frac{1}{5^2}).(1-\frac{1}{7^2})... $$ $$ P(Relatively Prime) = \frac{1}{\zeta(2)} $$ $$ P(Relatively Prime) = \frac{1}{\frac{\pi^2}{6}} $$ $$ P(Relatively Prime) = \frac{6}{\pi^2} $$

There we go!! Probability that any two numbers are relatively prime to each other is not a very complicated expression.

*[July 04 2019, Patan Dhoka]*