A is set containing n elements. A subset P of A is chosen at random. The set A reconstructed by replacing the elements of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements.

Set A has a elements.

∴ Number of subsets of A = 2”

∴ Each one of P and Q can be selected in 2” ways.

Hence total no. of ways of selecting P and Q = 2” = 4”.

Let P contains r elements, where r varies from 0 to n, Then, P can be chosen in ⁿ C_r ways.

Now as P ∩ Q = ϕ, Q can be chosen from the set of all subsets of set consisting of remaining (n – r) elements. This can be done in 2^{n – r} ways.

∴ P and Q can be chosen in ⁿ C_r­. 2^{n – r} ways. But, r can vary from 0 to n

∴ total number of disjoint sets P and Q are

= = (1 + 2)ⁿ = 3ⁿ

NOTE THIS STEP:

∴ Required probability = 3ⁿ/4ⁿ = (3/4)ⁿ

ALTERNATE SOLUTION:

Let A = {a₁, a₂, a₃, . . . . . . . . . . ., a_n}

For each a_i, 1 ≤ i n, there Aries 4 cases

(i) a₁ ∈ P and a₁ ∈ Q

(ii) a₁ ∉ P and a₁­ ∈ Q

(iii) a₁ ∈ P and a₁ ∉ Q

(iv) a₁ ∉ P and a₁ ∉ Q

∴ total no. of ways of choosing P and Q is 4ⁿ. Here case (i) is not favorable as P ∩ Q = ϕ

∴ For each element there are 3 favorable cases and hence total no. of favorable cases 3”

Hence prob. (P ∩ Q) = ϕ) = 3ⁿ/4ⁿ = (3/4)ⁿ

ALTERNATE SOLUTION:

The set P be the empty set, or one element set or two elements set . . . . . . . . . or n elements set. Then the set Q will be chosen from amongst the remaining n elements or (n – 1). Element for (n – 2) elements . . . . . . . . . . . . . . . or no elements. Now if P is the empty set then prob. of its choosing is ⁿC₀/2ⁿ, if it is one element set the then prob. of its choosing is ⁿ C₁/2ⁿ, and so on. When the set P consisting of r elements is chosen from A, then the prob. of choosing the set Q from amongst the remaining n - r elements 2^{n - r} /2ⁿ. Hence the prob. that P and Q have no common elements is given by

/2ⁿ . 2^{n – r}/2ⁿ = 1/4ⁿ 2^{n -r}

= 1/4ⁿ (1 + 2)ⁿ (Using Binomial thm.) = 3ⁿ/4ⁿ = (3/4)ⁿ