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

Explanation

Out of 3n consecutive integers, n integers will be of 3k type, n integers will be of 3k + 1 type and n integers will be of 3k + 2 type.
If sum of three integers selected is divisible by 3, then we have following possibilities.
All integers are of same type.
`therefore` Number of ways of selection = `.^(n)C_(3)+.^(n)C_(3) + .^(n)C_(3) = 3 xx .^(n)C_(3)`
One integer of each type.
`therefore` Number of ways of selection = `.^(n)C_(1)xx.^(n)C_(1) xx .^(n)C_(1) = n^(3)`
So, number of favourable ways = `3 xx .^(n)C_(3) + n^(3)`
Total number of ways = `.^(3n)C_(3)`
`therefore` Required probability = `(3xx .^(n)C_(3) + n^(3))/(.^(3n)C_(3)) = (3n^(2) - 3n + 2)/((3n-1)(3n-2))`