Question

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 Select the correct answer from above options

Answer 1

Correct Answer - D Since, three distinct numbers are to be selected from first 100 natural numbers. ⇒ n(S)=100C3 Efavourable events= All three of them are di visible by both 2 and 3 . ⇒ Divisible by 6 i.e. {6, 12, 18, ... , 96} Thus, out of 16 we have to select 3. ∴ Reqmred probability = 16C3 100C3 = 4 1155