Correct Answer - D
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.
Suppose P contains r elements. For P∪Q to be equal to A, we must choose Q in such a way that it contains remaining (n-r) elements and any number of elements from r elements chosen in P.
So, total number of ways of selecting P and Q is .nCr.2r
But, r can vary from 0 to n. Therefore,
Number of ways of selecting P and Q
=
n
∑
r=0
.nCr.2r=(1+2)n=3n
Hence, required probability =
3n
4n
=(
3
4
)n
ALITER If P∪Q=A, then each element must be listed in any one of three ways (i), (iii) and (iv) discussed in Example 42.
∴ Favourable number of elementary events =3n
Hence, required probability =
3n
4n