Out of `3n` consecutive integers, there are selected at random. Find the probability that their sum is divisible by 3.

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

Let the sequence of 3n consecutive integers begins with the integer m. Then, the 3n consecutive integers are
m,m+1,m+2,.., m+(3n-1).
Out of these integers, 3 integers can be chosen in `.^(3n)C_(3)` ways.
Let us divide these 3n consecutive integers into three groups `G_(1),G_(2) " and " G_(3)` as follows :
`G_(1) : m, m+3, m+6,.., m+(3n-3)`
`G_(2) : m+1, m+4, m+7,.., m+(3n-2)`
`G_(3) : m+2, m+5, m+8,.., m+(3n-1)`
The sum of 3 integers chosen from the given 3n integers will be divisible by 3 if either all the three integers are chosen from the same group or one integer is chosen from each group. The number of ways that the three integers are from the same group is `(.^(n)C_(3)+ .^(n)C_(3)+ .^(n)C_(3))` and the number of ways that the integers are form different groups is `(.^(n)C_(1)xx .^(n)C_(1)xx .^(n)C_(1))`.
So, the number of ways in which the sum of three integers is divisible by 3 is
`(.^(n)C_(3)+ .^(n)C_(3)+ .^(n)C_(3))+( .^(n)C_(1)xx .^(n)C_(1)xx .^(n)C_(1))=3 .^(n)C_(3)+( .^(n)C_(1))^(3)`
Hence, required probability `=(3xx.^(n)C_(3)+(.^(n)C_(1))^(3))/(.^(3n)C_(3))`
`=(3n^(2)-3n+2)/((3n-1)(3n-2))`