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If three distinct number are chosen randomly from the first 100 naturalnumbers, then the probability that all three of them are divisible by both 2and 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)` |
Answer» Correct Answer - 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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