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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

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$\frac{4}{25}$

$\frac{4}{35}$

$\frac{4}{33}$

$\frac{4}{1115}$

answer is D.

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Detailed Solution

If a number is to be divisible by both 2 and 3, it should be divisible by their L.C.M. L.C.M. of 2 and 3 is 6. The numbers are 6, 12, 18, .. .,96. The total number is 16. Hence, the probability is

$\frac{^{16} C_{3}}{^{100} C_{3}} = \frac{4}{1155}$

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