13 Dec
2004
13 Dec
'04
7:42 a.m.
<< Am I making some mistake here: Prob(p divides all three of K, L, M)=1/p^3 Prob(p does not divide all three) = 1 - 1/p^3 giving the final probability as the product of these terms over all p, which is 1/zeta(3) = 0.8319...
I don't understand the reason you choose to multiply those particular factors (for each prime p). The criterion is that either 0 or 1 of K,L,M can be divisible by p, but no more than 1 of them. --Dan Asimov