Let S be the universal set and (n) X = n . The probability of selecting two subsets A and B of the X such that `B = barA` is :

Options

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

Answer

B

Explanation

The set A has `2^(n)` elements. Therefore, two subsets P and Q can be chosen in `2^(n) C_(2)` ways.
Suppose P consists of r elements. Then, P can be chosen in `.^(n)C_(r )`.
Since, `Q=overline(P)`. Therefore, P and Q can be chosen in `.^(n)C_(r )` ways. But, r can vary from 0 to n and P and Q can be interchanged also.
`therefore` Number of ways of selecting P and Q such that `Q=overline(P)` is
`(1)/(2) underset(r=0)overset(n)sum .^(n)C_(r )=2^(n-1)`
Hence, required probability `=(2^(n-1))/(.^(2n)C_(2))=(1)/(2^(n)-1)`