Probabilities in a Set of numbers Let $S = \{1, 2, 3, \dots, n\}.$ Three subsets $A,$ $B,$ $C$ of $S$ are chosen at random.
(a) Find the probability that $A \cup B \cup C = S.$
(b) Find the probability that $A \subseteq B \subseteq C.$
Here is what I have for Part (a) so far:
For the equality to be false, there must be some element in S that is not in A or B or C. And for each element, there is a probability of 1/2 that it is in a given subset. It can be put into the subset, or it can be left out. So, for each element, there is a 1/8 chance it is not in any of the 3 subsets.
I'm stuck on how to continue and I have no idea how to do part b.
 A: You've made a good start on (a)- every element has a $\frac78$ chance of being covered. These events are independent so the probability every element is covered is $(\frac78)^n$.
You can do (b) similarly. In order to satisfy the condition, each element needs to be either in none of the sets, in $C$ only, in $B$ and $C$, or in all of them. What is the probability of this for each element?
A: Answer for part $b$. (SPOILERS)
To choose $A,B,C \subseteq S$ such that $A \subseteq B \subseteq C$, you can use this method:
-choose $c \in [\![0,n]\!]$ the cardinal of $C$
-choose the elements of $C$
-choose $b\in [\![0,c]\!]$ the cardinal of $B$
-choose the elements of $B$
-choose $a\in [\![0,b]\!]$ the cardinal of $A$
-choose the elements of $A$
And this uniquely determines $A,B,C$. Therefore, the total number of triplets $(A,B,C)$ of subsets of $S$ verifying $A \subseteq B \subseteq C$ is
$$
\sum_{c=0}^n  {n\choose{c}} \sum_{b=0}^c  {c\choose{b}} \sum_{a=0}^n  {b\choose{a}}
= \sum_{c=0}^n  {n\choose{c}} \sum_{b=0}^c  {c\choose{b}} 2^b
= \sum_{c=0}^n  {n\choose{c}} \underbrace{\sum_{b=0}^c  {c\choose{b}} 2^b}_{(1+2)^c = 3^c}
= \sum_{c=0}^n  {n\choose{c}} 3^b = 4^n
$$
