A is a set containing n elements, A subset P (may be void also) is selected at random from set A and the set A is then reconstructed by replacing the elements of P. A subset Q (may be void also) of A is again chosen at random. The probability that

Options

A. `((1)/(2))^(n)`
B. `((1)/(4))^(n)`
C. `(3)/(4)`
D. `((3)/(4))^(n)`

Answer

D

Explanation

The set A has n elements. So, it has `2^(n)` subsets.
Therefore, set P can be chosen in `2^(n)C_(1)` ways. Similarly, set Q can also be chosen in `2^(n) C_(1)` ways.
`therefore` Sets P and Q can be chosen in `.^(2n)C_(1)xx .^(2n)C_(1)=2^(n)xx2^(n)=4^(n)` ways.
Suppose P contains r elements, where r varies from 0 to n.
Then, P can be chosen in `.^(n)C_(r )` ways.
For Q to be disjoint from A, it should 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. Therefore, P and Q can be chosen in `.^(n)C_(r )xx2^(n-r)` ways.
But, r can vary from 0 to n. Therefore, the total number of ways of selecting P and Q such that they are disjoint is
`underset(r=0)overset(n)sum .^(n)C_(r ) 2^(n-r)=(1+2)^(n)=3^(n)`
Hence, required probability `=(3^(n))/(4^(n))=((3)/(4))^(n)`