Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now onemore subset B of P is drawn. Probability of drawing sets A and B so that A∩B has exactly one element is

Let xi be any element of set P, we have following possibilities1 xi∈A,xi∈B2 xi∈A,xi∉B3 xi∉A,xi∈B4 xi∉A,xi∉BClearly, the element    A o B it it belongs to A and B both.Thus out of these 4 ways only first way is favorable. Now the element that we want to be in the intersection can be chosenin'n' different ways. Hence required probability is n⋅(3/4)n−1