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 [2004]

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 [2004]

Smart Achievers · Accepted Answer

(4)

Total numbers which are divisible by both 2 and 3 i.e. 6 are 16

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

Solution

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 [2004]

1 4/25

2 4/35

3 4/33

4 4/1155

Ans.
(4)

Total numbers which are divisible by both 2 and 3 i.e. 6 are 16

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

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