If three numbers are selected from the set of the first 20 natural numbers, the probability that they are in GP, is

n(S) =²⁰ C3 = 1140
Three numbers are in GP, the favourable cases are
1,2,4; 1,3,9; 1,4,16; 2,4,8; 2,6,18; 3,6,12; 4,8,16;
5,10,20; 4,6,9; 8,12,18; 9,12,16
n(E) = 11
$\mathrm{P}\left(\mathrm{E}\right)=\frac{11}{1140}$