Question

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 Q is a subset of P, is
A. `(3)/(4^(n))`
B. `((3)/(4))^(n)`
C. `n((3)/(4))^(n)`
D. `(3n)/(4^(n))`

Answer 1

Correct Answer - B
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 )`. The subset Q of P can have at most r elements and the number of ways of choosing Q is `2^(r )`. Therefore, the number of ways of choosing P and Q is `.^(n)C_(r ) xx 2^(r )` when P has r elements. So, P and Q can be chosen in general in
`underset(r=0)overset(n)sum .^(n)C_(r )xx 2^(r )=(1+2)^(n)=3^(n)` ways
Hence, required probability `=(3^(n))/(4^(n))=((3)/(4))^(n)`