Probability Question 382
Question: 3 integers are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3, is.
Options:
A) $ \frac{194}{285} $
B) $ \frac{1}{57} $
C) $ \frac{13}{19} $
D) $ \frac{3}{4} $
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Answer:
Correct Answer: A
Solution:
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} $
BETA
