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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

  1. A4 35
  2. B4 55
  3. C4 1155
  4. 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

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