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

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 (A) 4 / 25 (B) 4 / 35 (C) 4 / 33 (D) 4 / 1155

Tardigrade Admin · Accepted Answer

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, ldots, 96. The total number is 16 . Hence; the probability is ( 16 C3/ 100 C3)=(4/1155)

Q. 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

$4/25$

$4/35$

$4/33$

$4/1155$

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 $6, 12, 18, \dots, 96$ . The total number is $16$ . Hence; the probability is
$\frac{^{16} C _{3}}{^{100} C _{3}} = \frac{4}{1155}$

Q. 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

4/25

4/35

4/33

4/1155