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

Answer

A

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.
The total number of ways of selecting P and Q such that P=Q, is
`underset(r=0)overset(n)sum .^(n)C_(r )=2^(n)`
Hence, required probability `=(2^(n))/(4^(n))=(1)/(2^(n))`
ALITER If P=Q, then each element can be listed in two ways (i), (ii) only.
`therefore` Favourable number of elementary events `=2^(n)`
Hence, required probability `=(2^(n))/(4^(n))`