Two number `a and b` are chosen at random from the set of first 30 natural numbers. Find the probability that `a^2-b^2` is divisible by 3.

Answer

`(47)/(87)`

Explanation

The total number of ways of choosing two numbers out of 1, 2, 3, …, 30 is `.^(30)C_(2)` = 435.
Now, 30 natural numbers are categorized as
3k type: 3, 6, 9, … 30 `rarr` 10 numbers
3k + 1 type: 1, 4, 7, …, 28 `rarr` 10 numbers
3k + 2 type: 2, 5, 8, ..., 29 `rarr` 10 numbers
`a^(2) - b^(2)` is divisible by 3 iff either both of a and b are So, number of favourable cases = `.^(10)C_(2) + .^(20)C_(2) = 235`
`therefore` Required probability = `(235)/(435) = (47)/(87)`