Correct Answer - B
The set A has n elements. So, it has
2
n
2
subsets.
Therefore, set P can be chosen in
2
n
C
1
2
ways. Similarly, set Q can also be chosen in
2
n
C
1
2
ways.
∴
∴
Sets P and Q can be chosen in
.
2
n
C
1
×
.
2
n
C
1
=
2
n
×
2
n
=
4
n
.
ways.
If
P
∪
Q
P
contains exactly one element, P and Q each can have at most one element. That is, if P has no element, Q must have one element, and the number of ways of choosing P and Q is
.
n
C
0
×
.
n
C
1
=
n
.
. On the other hand, if P has one element, Q can be empty or it can be equal to P i.e. Q can be chosen in two ways, so that the number of ways of choosing P and Q is
.
n
C
1
×
2
=
2
n
.
.
Therefore, the number of ways of choosing P and Q in this case is n+2n=3n.
Hence, the required probability
=
3
n
4
n
=