[bigpooch]

That's why I like Siegmund's solution. It follows quickly
that the following all have probability (3/4)^n:

P(A is a subset of B)
P(B is a subset of A)
P(A intersect B is the null set)
P(A union B is S, the "universe").

Also, these have probability (1/2)^n:

P(A=B)
P(A and B are complementary).

In addition, by looking at elements only, not only can you
generalize to more sets (say A, B and C are "random" subsets
of S; then, the probability[A is a subset of B and B is a
subset of C] = (1/2)^n), but you can also generalize the
conditions for which an element is in A or B (perhaps one
wants P(x is in A)=a and P(x is in B)=b) and solve such a
problem relatively easily using the "element approach".

Proposition: If S has n elements and A[1], A[2], ..., A[k]
are "independent uniform random subsets" of S, the
probability that A[j] is a subset of A[j+1] for all integers
j such that 1<=j<=(k-1) is then ((k+1)/(2^k))^n.