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?

Pick any subset $A$, and there is only one subset $B$, namely $S \setminus A$  which satisfies $A \cup B = S$ and $A \cap B = \emptyset $.
There are $2^n$ subsets to choose from so the probability of selecting such a pair is $1/2^n$.
(Or, $1/(2^n - 1)$ if one constrains that $A \ne B$).
