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))`

Answer

C

Explanation

In 3n consecutive natural numbers,
(i) n number are of form 3p.
(ii) n numbers are of form 3p + 1.
(iii) n numbers are of form 3p + 2.
Here, favourable number of cases = Either we can select three numbers from any of the set or we can select one from each set
`= .^(n)C_(3)+ .^(n)C_(3)+ .^(n)C_(3)+(.^(n)C_(1)xx .^(n)C_(1)xx .^(n)C_(1))`
`= 3((n(n-1)(n-2))/(6))+ n^(3)=(n(n-1)(n-2))/(2)+n^(3)`
Total number of selections `= .^(3n)C_(3)`
`therefore` Required probability `=((n(n-1)(n-2))/(2)+n^(3))/((3n(3n-1)(3n-2))/(6))`
`=(3n^(2)-3n+2)/((3n-1)(3n-2))`