If three distinct number 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 `4//25` b. `4//35` c. `4//33` d. `4//1155`

Options

A. `(4)/(25)`
B. `(4)/(35)`
C. `(4)/(33)`
D. `(4)/(1155)`

Answer

D

Explanation

A number will be divisible by both 2 and 3, if it is divisible by their 1.c.m. i.e. 6.
There are 16 numbers, in first 100 natural numbers, which are divisible by 6. Therefore, number of ways of selecting 3 numbers such that all of them are divisible by both 2 and 3 is `.^(16)_(3)`.
`therefore` Required probability `=(.^(16)C_(3))/(.^(100)C_(3))=(16xx15xx14)/(100xx99xx98)=(4)/(1155)`