Show $P(A \subset B) = (\frac{3}{4})^n$ ​Let $S = \{1, 2, \dots, n \}$ and​​ suppose that $A$ and $B$ are,
  independently, equally likely to be any of the $2^n$ subsets of $S$.
  
  Show that $P(A \subset B) = (\frac{3}{4})^n$ and show that $P(A \cap B = \emptyset) = (\frac{3}{4})^n$.


Applying the law of total probability ,

$$P(A \subset B)  =\sum_{k=0}^n P(A \subset B \Large \mid \normalsize |B| = k)P(|B| = k) $$

If $k = 0$ then $B = \varnothing $. Therefore  $P(A \subset B \Large \mid \normalsize |B| = k) = 0$ because there do not exist any sets $A$ such that $A \subset B$. 

If $k = 1$ and $A \subset B$, then there exists precisely one set $A$ such that $A \subset B$, and this $A = \varnothing$. Therefore  

$$\begin{align}
&amp;P(A \subset B \mid \normalsize |B| = k) =  \frac{1}{2^n} = \frac{2^1 - 1}{2^n} = \frac{2^k - 1}{2^n} \\ 
&amp;P(|B| = k) = \frac{\binom{n}{k}}{2^n}
\end{align}$$

If  $A \subset B$, then $A \in  \mathcal{P}(B) - B$. Thus, there are $|\mathcal{P}(B) - B| = 2^{k} - 1$ possibilities for $A \subset B$ out of a possible $2^{|S|} = 2^n$ sample space. 

Therefore I have 

$$\begin{align}
&amp;P(A \subset B)  = 0 + \sum_{k=1}^n P(A \subset B \Large \mid \normalsize |B| = k)P(|B| = k) = \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n}  \\
&amp;\mathsf{but} \ \ \  \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n} \ne (\frac{3}{4})^n
\end{align}$$

Where is my mistake? Hints only please.
It is easy to see that $P(A\subseteq B)=(\frac34)^n$ so the first part of what you're trying to show can't be true.

Namely, suppose we choose for each of the $n$ possible elements first whether it is in $A$ and then whether it is in $B$. All of the $2n$ choices are independent and uniformly distributed between yes and no.

$A\subseteq B$ means that each pair of answers is one of (yes, yes), (no, yes) or (no, no). For each pair the probability of that happening is $\frac 34$, so the probabilty of that happening for all $n$ pairs is $(\frac{3}{4})^n$.

Similarly $A\cap B=\varnothing$ means that each pair of answers is one of (yes,no), (no, yes), or (no,no). Again, this has probability $\frac 34$ and must happen $n$ times, so the probability of disjoint $A$ and $B$ is $(\frac 34)^n$.



Trying to count possibilities, grouped by the size of the sets is definitely the long way around to an answer here.
If you want to calculate strictly about the case of $A\subset B$, from my persepective there is nothing wrong about your calculation. But if you want to consider about the case of $A\subseteq B$， the above answer is direct and simple. I  would like to give a solution using the law of total probability.
$$
P(A \subseteq B)=\sum\limits_{i=0}^{n} P(A\subseteq B \mid |B|=i)\cdot P(|B|=i)=\sum\limits_{i=0}^{n} \frac{2^i}{2^n} \cdot \frac{\left(\begin{array}{c}
n\\i\end{array}\right)}{2^n}=\frac{\sum\limits_{i=0}^{n}\left(\begin{array}{c}
n\\i\end{array}\right)2^i1^{n-i}}{4^n}=\frac{3^n}{4^n}
$$
The second question is similar.