Let A={a1,a2,...,an}. For each a1ϵA(1≤i≤n) we have
the following four cases:(i) a1ϵP
and a1ϵQ
(ii) a1/ϵP and a1ϵQ
(iii) a1ϵP and a1/ϵQ
(iv) a1/ϵP and a1/ϵQ Thus, the total number of ways of choosing P and Q is 4n.
A) Out of these four
choices (i) is not favourable for P∩Q=ϕ. Thus, probability thatP∩Q is (3/4)n.
B) P∩Q can
contain exactly one element in nC1(3n−1)=n(3n−1) ways.
∴
Probability that P∩Q is singleton =(3n−1/4n
C) P∪Q=A in
3n ways.
∴
Probability P∪Q=A is (3/4)n.
D) |P|=|Q| in the following ways.
(nC0)2+(nC1)2+....+(nCn)2=2nCn.