NIMCET 2024 Paper

Section: Mathematics

Question : 48 of 120

Marks: +1, -0

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

$\;{4}/{33}$
$\;{4}/{35}$
$\;{4}/{25}$
$\;{4}/{1155}$

Validate

Solution:

All the three numbers should be multiple of 6 , as there are 16 multiples of 6 in the first 100 natural numbers, hence the required probability

=\;{{ }^{16} C_3}/{{ }^{100} C_3}=\;{16 × 15 × 14}/{100 × 99 × 98}=\;{4}/{1155}

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