A) \[\frac{2}{19}\]
B) \[\frac{3}{29}\]
C) \[\frac{17}{19}\]
D) \[\frac{4}{19}\]
Correct Answer: C
Solution :
The total number of ways in which 3 integers can be chosen from first 20 integers is \[^{20}{{C}_{3}}.\] The product of three integers will be even if at least one of them is even. \[\therefore \]Required probability = 1 ? Probability that none is even \[=1-\frac{^{10}{{C}_{3}}}{^{20}{{C}_{3}}}=1-\frac{2}{19}=\frac{17}{19}.\]
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