An interesting (and famous) probability

A randomly chosen natural nonzero number $n$ evenly divides every $n$ th integer. The probability, then, that a natural nonzero number divides a random integer is $\frac {1}{n}$ .

Every integer can be written as a product of prime numbers, so two coprime integers can be considered as numbers which share no prime numbers in their factorization.

Now we combine these two realizations together!

The probability of two integers being divisible by the same integer $a$ is simply $\frac {1}{a^2}$ . So the sum of this probability for each prime number will give us our desired result, the probability that two random positive integers are coprime.

This can be written as $\prod _{ n=1 }^{ \infty }{ \left( 1-\frac { 1 }{ { P }_{ n } } \right) }$ where ${P}_{n}$ is the $n$ th prime starting at $2$ . This is rewritten as $\prod _{ n=1 }^{ \infty }{ \left( \frac { { P }_{ n }-1 }{ { P }_{ n } } \right) }$ which is the inverse of one of our given products.

It follows that the probability (chaining into the given sum of $\frac {1}{n^2}$ ) is $\frac{6}{\pi^2}$ , which, when rounded to 3 decimal places, is $\boxed{0.608}$