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If 3 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

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$\frac{4}{55}$

$\frac{4}{35}$

$\frac{4}{33}$

$\frac{4}{1155}$

answer is D.

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Detailed Solution

The number to be selected are from the set ${6, 12, 18, . . ., 96}$ of 16 numbers.

$∴$ Required probability $= \frac{(16_{C_{3}})}{(100_{C_{3}})}$

$= \frac{16 \cdot 15 \cdot 14}{100 \cdot 99 \cdot 98} = \frac{4}{1155}$

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