Two natural numbers x and y are chosen at random from the set `{1,2,3,4,...3n}`. find the probability that `x^2-y^2` is divisible by 3.

Explanation

Let S = Sample space
and E = the event that `x^2-y^2` is divisible by 3.
n(S) = number of ways of choosing 2 numbers out of 3n number
`=""^(3n)C_2=(3n(3n-1))/(2)`
Partition the set {1, 2, 3,…. 3n} into three mutually disjoint sets
`A_0`={3, 6, 9, .....3n}
`A_1`={1, 4, 7, 10, ..... 3n-2}
`A_2`={5, 8, 11, ..... 3n-1}
`A_0` consists of all numbers of the form 3k, `A_1` of form 3k+1, and `A_2` of the form 3k+2.
Now `x^2-y^2` will be divisible by 3 if (x-y)(x+y)is divisible by 3. For this to happen, either both x and y should come from `A_0 " or " A_1 " or " A_2` or one is selected from `A_1` and the other form `A_2`.