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AIEEE 2004

JEE Main 2004 Mathematics question on Probability

If three distinct numbers 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
  1. A.
    455\dfrac{4}{55}
  2. B.
    435\dfrac{4}{35}
  3. C.
    433\dfrac{4}{33}
  4. D.
    41155\dfrac{4}{1155}
Show answer and solution

Correct option: D

Solution

Numbers divisible by both 2 and 3 must be divisible by 6. Multiples of 6 from 1 to 100: 6,12,...,96 → 16 numbers.

P = C(16,3)/C(100,3) = (16×15×14)/(100×99×98) = 3360/970200 = 4/1155.

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