Theorems of probability

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Question

Three integers are chosen from 1 to 20 without replacement. Probability that their product is even, is

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Solution

Step-I:
$Out of 20 integers, 3 can be selected in {}^{20}C_{3} ways$
$to be product even$ at least one integer is even
= Total-odd
$= 20_{C_{3}} - 10_{C_{3}}$
$R \cdot P = \frac{20_{C_{3}} - 10_{C_{3}}}{20_{C_{3}}} = 1 - \frac{10_{C_{3}}}{20_{C_{3}}}$
$= \frac{17}{19} = 0.89$

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