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

Options

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

Explanation

Let the sequence of 3n consecutive integras begins with m. Then 3n consective integers are `m, m+1, m+2, .....m+(3n+1)`
3 integers from 3n can be selected in `.^(3n)C_(3)` ways
`:.` Total number of outcomes `=.^(3n)C_(3)`
Now 3n integers can be divided into 3 groups.
`G_(1)` : n numbers of from 3p
`G_(2)` : n numbers of form 3p+1
`G_(3)` : n number of form 3p+2
The sum of 3 integers chosen from 3n integers will be divisible by 3 if either all the three are from same group of one integer from each group. The number of ways that the three integers are from same group is `.^(n)C_(3)+.^(n)C_(3)+.^(n)C_(3)` and number of ways that the integers are from different group is `.^(n)C_(1)xx.^(n)C_(1)xx.^(n)C_(1)`
`:.` Favourable cases `=(.^(n)C_(3)+.^(n)C_(3)+.^(n)C_(3))+(.^(n)C_(1)xx.^(n)C_(1)xx.^(n)C_(1))`
`:.` Required probability = `(3.^(n)C_(3)+(.^(n)C_(1))^(3))/(.^(3n)C_(3))`
`=(3n^(2)-3n+2)/((3n-1)(3n-2))`