Correct Answer - B
The set A has n elements. So, it has 2n subsets.
Therefore, set P can be chosen in 2nC1 ways. Similarly, set Q can also be chosen in 2nC1 ways.
∴ Sets P and Q can be chosen in .2nC1×.2nC1=2n×2n=4n ways.
Let the subset P of A contains r elements, with 0≤r≤n. Then, the number of ways of choosing P is .nCr. The subset Q of P can have at most r elements and the number of ways of choosing Q is 2r. Therefore, the number of ways of choosing P and Q is .nCr×2r when P has r elements. So, P and Q can be chosen in general in
n
∑
r=0
.nCr×2r=(1+2)n=3n ways
Hence, required probability =
3n
4n
=(
3
4
)n