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If three distinct number are chosen rand...

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`

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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