Question

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

• A. (3/4)ⁿ.n
• B. n.(3/4)^{n – 1}
• C. n.(3/4)ⁿ
• D. None

Answer 1

Answer: (B)

n.(3/4)^{n – 1}

Answer 2

Let x_i be any element of set P, we have following

possibilities

(i) x_i Ī A, x_i Ī B ; (ii) x_i Ī A, x_i Ļ B ;

(iii) x_i Ļ A, x_i Ī B ; (iv) x_i Ļ A, x_i Ļ B

Clearly, the element x_i Ī A Ē B if 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 chosen in 'n' different ways. Hence required probability is n.(3/4)^{n – 1}.