Three distinct numbers are selected from 100 natural number. The probability that all the three numbers are divisible by 2 and 3 is

$\frac{4}{1155}$

The numbers should be divisible by 6. Thus the number of favourable ways is ${^{16}C}_3$ (as there are 16 numbers in first 100 natural numbers, divisible by 6). Required probability is

$\frac{^{16}C_3}{^{100}C_3}=\frac{16×15×14}{100×99×98}=\frac{4}{1155}.$