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?
 A: 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 $A \ne B$).
A: Assuming you mean that $A$ and $B$ are picked independently and uniformly at random.
Edit: below, I'm answering two different questions. I guess, after reading yours carefully, that what you intended is the second one.
First interpretation of the question: "find $\Pr[A\cap B=\emptyset]$ and  $\Pr[A\cup B=S]$"
Consider a fixed element $s\in S$. The probability that it belongs to neither $A$ nor $B$ is
$$
\Pr[s\notin A \text{ and } s\notin B] = \Pr[s\notin A]\cdot \Pr[s\notin B] = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}
$$
where we used independence for the first equality. We have $A\cup B = S$ if, and only if, all elements belong to at least one of $A$ and $B$, which by the above happens for each fixed element $s$ with probability $1-\frac{1}{4}$. Again by independence, this results in
$$
\Pr[ A\cup B= S] = \Pr[\forall s,\ s\in A\cup B]
= \prod_{s\in S} \Pr[s\in A\cup B] = \left(\frac{3}{4}\right)^n.
$$
Similarly, the probability that a given fixed element $s\in S$ belongs to $A\cap B$ can be shown to be $\frac{1}{4}$. The probability that no element ends in $A\cap B$ is then
$$
\Pr[ A\cap B=\emptyset] = \prod_{s\in S}\Pr[ s\notin A\cap B]
= \prod_{s\in S}\left(1-\Pr[ s\in A\cap B]\right) = \left(1-\frac{1}{4}\right)^n=\left(\frac{3}{4}\right)^n.
$$
Second interpretation: "find $\Pr[A\cap B=\emptyset \text{ and } A\cup B=S]$"
This is similar to the above approach. Fix any element $s\in S$: the probability that $s$ ends up in exactly one of $A$ and $B$ is 
$$\begin{align}
\Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]
&=\Pr[s\in A \text{ and } s\notin B] + \Pr[s\notin A \text{ and } s\in B]\\
&=\Pr[s\in A]\cdot\Pr[s\notin B] + \Pr[s\notin A]\cdot\Pr[s\in B]\\
&=\frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}
=\frac{1}{2}
\end{align}$$
so the probability you want is
$$\begin{align}
\Pr[ \forall s\in S,\ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]
&=\prod_{s\in S }\Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]\\
&=\left(\frac{1}{2}\right)^n
\end{align}$$
A: 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.
A: Let $\Sigma = \{(A,B) | A,B \subset S \}$, we see that $|\Sigma| = 2^n 2^n = 4^n$. I am assuming that each pair $(A,B)$ is equipropable.
Let $C_1 = \{(A,B)| A \cup B = S \}$.
The sets $E_A = \{(A,B)| A \cup B = S \} = \{(A,B)| A^c \subset B \}$
form a partition of $C_1$ and we see that $|E_A| = 2^{|A|}$, since $B$
must be of the form $B=A^c \cup D$, where $D \subset A$, and the number of
such $D$ is $2^{|A|}$.
Hence $|C_1| = \sum_{A \subset S}  2^{|A|} = \sum_{k=0}^n \binom{n}{k} 2^k = 3^n$, and hence $p C_1 = ({3 \over 4})^n$
Let $C_2 = \{(A,B)| A \cap B = \emptyset \} $.
The sets $F_A = \{(A,B)| A \cap B = \emptyset \} = \{(A,B)|  B\subset A^c \}$ form a partition of $C_2$ and we see that $|F_A| = 2^{n-|A|}$, since $|A^c| = n-|A|$.
Hence $|C_2| = \sum_{A \subset S}  2^{n-|A|} = \sum_{k=0}^n \binom{n}{k} 2^{n-k} = 3^n$, and hence $p C_2 = ({3 \over 4})^n$
A: Right sorry I have a moment free again... The number of "successful" outcomes is the number of ways of composing two disjoint sets whose union is exhaustive over $S$.  The probability is that divided by the total number of ways of distributing the elements to A, B, A & B, or neither - which gives $4^n$ total possible outcomes.
A partition into two mutually disjoint sets is enumerated by the power set in selection of A, with a unique corresponding B for each A defined as the complement of A.
Therefore the answer is $2^n/4^n = 1/2^n$.
A: 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 $(1/4)^n+(1/4)^n$.
