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Out of 3n consecutive natural numbers, 3...

Out of 3n consecutive natural numbers, 3 natural numbers are chosen at random without replacement. The probability that the sum of the chosen numbers is divisible by 3, is

A

`(n(3n^2-3n+2))/2`

B

`((3n^2-3n+2))/(2(3n-1)(3n-2))`

C

`((3n^2-3n+2))/((3n-1)(3n-2))`

D

`(n(3n-1)(3n-2))/((3n-1))`

Text Solution

Verified by Experts

The correct Answer is:
C
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