Question

A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of the subset P. A subset Q of A is again chosen at random. The probability such that P ∩ Q contains 2 elements is (ⁿC_a.3^{n - b})/4ⁿ, then find a + b.

Answer 1

Let A = {a₁, a₂, …, a_n}. For each a_i ∈ A(1 ≤ i ≤ n), we have the following four cases ; 

(i) a_i ∈ P and a_i ∈ Q 

(ii) a_i ∉ P and a_i ∈ Q 

(iii) a_i ∈ P and a_i ∉ Q 

(iv) a_i ∉ P and a_i ∉ Q 

Thus, the total number of ways of choosing P and Q is 4ⁿ P ∩ Q contains exactly two element in (ⁿC₂) (3^{n – 2}).

Hence, the probability of P ∩ Q contains two elements is

(ⁿC₂.3^{n - 2})/4ⁿ

Here,

a = 2, b = 2 ⇒ a + b = 4