First slide

Theorems of probability

Question

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

Easy

A

$\frac{4}{25}$
B

$\frac{4}{35}$
C

$\frac{4}{33}$
D

$\frac{4}{1115}$

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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