Correct Answer - C::D
Since, set A contains n elements n elements. So, it has 2n subsets.
∴ Set P can be chosen in 2n ways, similarly set Q can be chosen in 2n ways.
∴ P and Q can be chosen in (2n)(2n)=4n ways.
Suppose, P contains r elements, where varies from 0 to n. Then, P can be chosen in
n
C
r ways, for 0 to be disjoint from A, it should be chosen from the set of all subsets of set consisting of remainning (n-r) elements. This can be done in 2n-r ways.
∴ P and Q can be chosen in
n
Cr.2n-r ways.
But, r can vary from 0 to n.
∴Totalνmberofdisj∮setsPandQ=overset(n)underset(r=0)sumoverset(n)""C_(r )2^(n-2)=(1+2)^(n)=3^(n)Hence,requiredprobability=(3^(n))/(4^(n))=((3)/(4))^(n)`