A) \[\frac{194}{285}\]
B) \[\frac{1}{57}\]
C) \[\frac{13}{19}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Total possible way\[n(S){{=}^{20}}{{C}_{3}}\] Let consider product is not a multiple of 3 then possible number of ways \[{{=}^{14}}{{C}_{3}}\]Hence, required probability\[=1\frac{^{14}{{C}_{3}}}{^{20}{{C}_{3}}}\] \[=1-\frac{91}{285}=\frac{194}{285}\]
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