Three integers are chosen at random from the set of first $20$ natural numbers. The chance that their product is a multiple of $3$ is

194 / 285

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1 / 57

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13 / 19

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3 / 4

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Solution

The correct option is A $194 / 285$
Total number of ways = ${^{20} C}_{3}$
It is a multiple of 3 when at least one of the three number is multiple of 3
Multiple of 3 are 3,6,9,12,15,18 are 6 nos.

Total favourable ways are by choosing any 1,2 or 3 in these 6 numbers
So total favourable ways = ${6_{C}}_{1} \times {14_{C}}_{2} + {6_{C}}_{2} \times {14_{C}}_{1} + {6_{C}}_{3} \times {14_{C}}_{0} = 776$
Probability $= \frac{776}{^{20} C_{3}} = \frac{194}{285}$

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Three integers are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3 is

Three integers are chosen at random from the set of first $20$ natural numbers. The chance that their product is a multiple of $3$ is