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Question

(C) The total number of ways to choose $3$ integers from the first $20$ integers is $^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1

Vedclass · Accepted Answer

$\frac{17}{19}$

Vedclass · Answer

(C) The total number of ways to choose $3$ integers from the first $20$ integers is $^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$.The product of three integers is even if at least one of the chosen integers is even.It is easier to calculate the probability of the complement event,which is that the product is odd. The product is odd if and only if all three chosen integers are odd.There are $10$ odd integers and $10$ even integers in the first $20$ integers.The number of ways to choose $3$ odd integers is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.The probability that the product is odd is $P(	ext{odd}) = \frac{^{10}C_3}{^{20}C_3} = \frac{120}{1140} = \frac{12}{114} = \frac{2}{19}$.Therefore,the probability that the product is even is $P(	ext{even}) = 1 - P(	ext{odd}) = 1 - \frac{2}{19} = \frac{17}{19}$.

Three integers are chosen at random from the first $20$ integers. The probability that their product is even is:

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