Probability that one randomly chosen subset of a finite set is a subset of another If $α$ and $β$ are subsets of $\{1,\dots,n\}$, what is the probability that $α \subseteq β$, given that $α$ and $β$ are chosen independently of each other and with the same probability.
I know that there are $2^n$ possible subsets of $\{1,\dots,n\}$ (including the empty set and the whole $\{1,\dots,n\}$) and if $|β| = k$ then there are $2^k$ possible subsets of $β$, which can be our $α$. 
I suspect (but I'm not 100% sure) that probability space $\Omega$ has a size of $2^{2n}$. If this is true I need to find the number of ways in which $α$ can be a subset of $β$. I'm not really sure how to do this since I don't know the number of elements in $β$. 
Can someone perhaps give me a little hint? 
I'm so sorry if this problem is too trivial, I'm very much a beginner and I find this to be pretty hard.
 A: The number of subsets of $\{1,...,n\}$ with $k$ elements is ${n \choose k}$ so, following your logic, the probability is $$\frac{1}{2^{2n}} \sum_{k=0}^n {n \choose k} 2^k.$$
Hint: You can simplify this to something related to $\frac34$ by considering the expansion of $(2+1)^n$.
A: Here is a different approach:
Take $A,B$ independent and uniformly random subsets of $\{1,..,n\}$. Since $|A\cup B|$ follows a Binomial$(n,3/4)$ distribution,${}^\dagger$ we have
$$
\Pr[A\cup B = \{1,...,n\}] = (3/4)^n. \tag{1}
$$
(the Binomial statement is even a bit overkill, but true).
Now, set $A=\{1,...,n\}\setminus \alpha$ and $B=\beta$. Then
$
\alpha\subseteq \beta$ if, and only if, $A\cup B = \{1,\dots,n\}$, so
$$
\Pr[\alpha\subseteq \beta] = \Pr[A\cup B = \{1,\dots,n\}] = (3/4)^n. \tag{2}
$$

${}^\dagger$ As every element $k$ independently belongs to $A\cup B$ with probability $1-\Pr[k\notin A]\Pr[k\notin B] = 1-1/2^2$.
A: We stop in front of each of $1,2,3,\dots,n$ and we have four equally likely choices: (i) in neither $\alpha$ nor $\beta$; (ii) in both; (iii) in $\beta$ but not in $\alpha$; in $\alpha$ but not in $\beta$.
For $\alpha\subseteq \beta$ we need have made one of the first $3$ choices $n$ times in a row.
