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 P. A subset Q is again chosen at random. The Probability that P and Q have equal number of elements, is

Options

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

Answer

B

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.
Let the subset P of A contains r elements, with `0 le r le n`. Then, the number of ways of choosing P is `.^(n)C_(r )`.
Similarly, Q can be chosen in `.^(n)C_(r )` ways.
So, P and Q can be chosen in `.^(n)C_(r )xx .^(n)C_(r )` ways. But, r can vary from 0 to n.
`therefore` P and Q can be chosen in `underset(r=0)overset(n)sum .^(n)C_(r )xx .^(n)C_(r )` ways `=.^(2n)C_(n)` ways.
`therefore` Required probability `=(.^(2n)C_(n))/(4^(n))`