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 cup Q` contain just one element, is

Options

A. `(3)/(4^(n))`
B. `(3n)/(4^(n))`
C. `n((3)/(4))^(n)`
D. `(n)/(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.
If `P cup Q` contains exactly one element, P and Q each can have at most one element. That is, if P has no element, Q must have one element, and the number of ways of choosing P and Q is `.^(n)C_(0)xx .^(n)C_(1)=n`. On the other hand, if P has one element, Q can be empty or it can be equal to P i.e. Q can be chosen in two ways, so that the number of ways of choosing P and Q is `.^(n)C_(1)xx2=2n`.
Therefore, the number of ways of choosing P and Q in this case is n+2n=3n.
Hence, the required probability `=(3n)/(4^(n))`