I want to solve the following exercise:
Suppose that two sets $X$ and $Y$ are chosen independently and uniformly at random from all the $2^n$ subsets of $\{1, \dotsc, n\}$.
Determine $P[X \subseteq Y]$ and $P[X \cup Y = \{1, \dotsc, n\}]$.
MY IDEA:
My idea is that a random subset is the same as deciding for each element independently with probability $\frac{1}{2}$ whether it is in the subset or not.
Then $$P[X \subseteq Y] = P[ \forall x \in X \colon x \in Y]$$
Now I have the first problem, because I am not sure how to express the for all $x$ that are in $X$. I mean, is this the same as computing $$\sum_{x=1}^n P[x \in X] P[x \in Y | x \in X] = \sum_{x=1}^n P[x \in X] P[x \in Y] = \frac{n}{4},$$ since this computation does not feel right. Can someone intuitively tell me what the above calculation calculates and how to solve the actual problem? My problem is here that we have somehow to condition on the event that this $x$ is already in $X$, otherwise we need not do anything.
The second one sounds a bit easier to me. There we have $$P[X \cup Y = \{1, \dotsc, n\}] = P[\forall x \in \{1, \dotsc, n\} \colon x \in X \vee x \in Y] =\prod_{x=1}^n P[x \in X \vee x \in Y] = \prod_{x=1}^n (P[x \in X] + P[x \in Y] - P[x \in X \wedge x \in Y] ) =\prod_{x=1}^n (P[x \in X] + P[x \in Y] - P[x \in X] P[ x \in Y] = \prod_{x=1}^n (\frac{1}{2}+\frac{1}{2}-\frac{1}{4})= \left(\frac{3}{4}\right)^n.$$ Is this correct? Is there an easier way?
For the first question, $P(X \subset Y)$, you've overcomplicated it a bit. To have $X \subset Y$ you just need that there are no values $k$ for which $k \in X$ and $k \not \in Y$. For any given $k$ the probability that $k \in X$ and $k \not \in Y$ is $3/4$, so the overall probability is $(3/4)^n$.
This is basically the same as your evaluation of $P(X \cup Y = \{ 1, 2, \ldots, n \})$ -- there are four cases, and in each problem one of them is bad.