A) \[\frac{194}{285}\]
B) \[\frac{1}{57}\]
C) \[\frac{13}{19}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
[a] Total number of ways of selecting 3 integers from 20 natural numbers \[{{=}^{20}}{{C}_{3}}=1140\] Their product is a multiple of 3 means, at least one number is divisible by 3. The numbers which are divisible by 3 are 3, 6, 9, 12, 15, 18 and the number of ways of selecting at least one of them is \[^{6}{{C}_{1}}{{\times }^{14}}{{C}_{2}}{{+}^{6}}{{C}_{2}}{{\times }^{14}}{{C}_{1}}{{+}^{6}}{{C}_{3}}=776\] Required Probability \[=\frac{776}{1140}=\frac{194}{285}\]
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