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 $A \ne B$).
Your proposed solution assumes A and B are disjoint in all possible cases.  The question appears to intend A and B are not necessarily disjoint, and requires them to be disjoint for success.

Also you don't state whether it's equally likely to choose e.g. 3 or 1 elements since the natural scenario in the real world is often a binomial distribution, with ~n/2 being the most likely number of elements chosen each time.  But let's assume that's not the case here.

Each element has 4 options: Only A, Only B, both or neither so there are $4^n$ cases.

The cases not fitting the first condition are those in which one element was distributed to neither, the probability being $1-(3/4)^n$.

The cases not fitting the 2nd condition are those in which one element was distributed to both, the probability again being $1-(3/4)^n$.

The combination of the two is a conditional probability so these cannot simply be multiplied.  The quickest way is to deduct the two probabilities of failure from 1 so the answer is $(1/4)^n+(1/4)^n$.