知道答案就簡單了.
The probability is (1-1/2^2)(1-1/3^2) ... (1-1/p^2)... because for different primes p,q, a number is a multiple of p is indepdent of it is a multiple of q.
On the other hand \pi^2/6 = \sum_i 1/n^2 = \prod_{p a prime} (1+1/p^2+1/p^4+...) = \prod_{p} 1/(1-1/p^2).
