If three numbers are choosen randomly from the first 100 natural numbers, then the probability that all the three of them are divisible by both 2 and 3 is

If a number is to be divisible by both 2 and 3, it should be divisible by their L.C.M,
L.C.M of 2 and 3 is 6
$\therefore$ The numbers are 6, 12, 18……96.

$\therefore$The total numbers are 16.

$\therefore$Required probability $=\frac{{16}_{C_3}}{{100}_{C_3}}=\frac{4}{1155}.$