If three distinct numbers are chosen randomly from the  first 100 natural numbers, then the probability that all  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. The numbers are 6, 12, 18, .. .,96. The total number is 16. Hence, the probability is  16C3 100C3=41155