Correct Answer - C
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, both P and Q must be non-empty. Thus, if P has r elements, Q must have exactly one of these r elements and any number of elements from among the remaining (n-r) elements in A, so that the number of ways of choosing Q is
.
r
C
1
×
2
n
−
r
.
But, P can be chosen in
.
n
C
r
.
ways. Therefore, P and Q can be chosen in
.
n
C
r
×
.
r
C
1
×
.2
n
−
r
.
, when P contains r elements.
Therefore, P and Q in general can be chosen in
n
∑
r
=
1
.
n
C
r
×
.
r
C
1
×
2
n
−
r
ways
=
n
∑
r
=
1
.
n
C
r
r
2
n
−
r
∑
ways
=
n
∑
r
=
1
n
r
.
n
−
1
C
r
−
1
.
r
.
2
n
−
r
[
∵
.
n
C
r
=
n
r
.
n
−
1
C
r
−
1
]
=
=
n
n
∑
r
=
1
.
n
−
1
C
r
−
1
2
n
−
r
=
=
n
n
∑
r
=
1
.
n
−
1
C
r
−
1
2
(
n
−
1
)
−
(
r
−
1
)
=
n
(
1
+
2
)
n
+
1
=
n
3
n
−
1
=
Hence, required probability
=
n
×
3
n
−
1
4
n
=
n
3
(
3
4
)
n
=
ALITER Let us first select an element out of n elements. This can be done in
.
n
C
1
.
ways. List this element in both P and Q.
Now, each of the remaining can be listed in three ways (ii), (iii) and (iv) discussed in example 42. This can be done in
3
n
−
1
3
ways.
∴
∴
Favourable number of elementary events
=
n
.
3
n
−
1
=
Hence, required probability
=
n
×
3
n
−
1
4
n
=
n
3
(
3
4
)
n
=