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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

A

`(4)/(55)`

B

`(4)/(35)`

C

`(4)/(33)`

D

`(4)/(1155)`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, three distinct numbers are to be selected from first 100 natural numbers. `rArr " " n(S)=""^(100)C_3`
`E_("favourable events")`= All three of them are di visible by both 2 and 3 .
`rArr ` Divisible by 6 i.e. {6, 12, 18, ... , 96} Thus, out of 16 we have to select 3.
`therefore` Reqmred probability `=(""^(16)C_3)/(""^(100)C_3)=4/1155`
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