MHT CET20225 Aug 2022Morning ShiftMathematicsProbabilityActual
If three distinct numbers are chosen randomly from first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is
Options
- A4 35
- B4 55
- C4 1155
- D80 231
Correct answer
C. 4 1155
Step-by-step solution
First 100 natural numbers are 1,2,3,4,5, . ., 100 Numbers divisible by both 2 and 3 are 6,12,18, . ., 96 (total 16) Now the required probability = ¹⁶ C₃ ¹⁰⁰ C₃ = 16 16-3 100 . . 13 100-3 aligned & = 16 3 3 13 3 97 100 & = 14 15 16 98 99 100 = 4 1155 & aligned