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
.
Similarly, Q can be chosen in .nCr
ways.
So, P and Q can be chosen in .nCr×.nCr
ways. But, r can vary from 0 to n.
∴
P and Q can be chosen in
n
∑
r=0
.nCr×.nCr
ways =.2nCn
ways.
∴
Required probability =
.2nCn
4n