Probability that $A \cup B$ = S and $A \cap B = \phi $ Let $S$ be a set containing $n$ elements and we select two subsets: $A$ and $B$ at random then the probability that $A \cup B$ = S and $A \cap B = \varnothing $ is?
My attempt
Total number of cases= $3^n$ as each element in set $S$ has three option: Go to $A$ or $B$ or to neither of $A$ or $B$
For favourable cases: Each element has two options: Either go to $A$ or to $B$ which gives $2^n$ favourable cases.
Is my approach correct?

The idea is good, but the total number of cases is not $3^n$. 
Let us assume that subsets are chosen independently, with all pairs of subsets equally likely. Then in effect for every element of $S$ we flip a fair coin twice. If the result is head, head, the element ends up in both $A$ and $B$. If the result is head, tail, then the element ends up in $A$ but not in $B$. And so on. So there are $4^n$ equally likely possibilities.
