I was working through this (Cambridge STEP 3 probability) question:
The set S is a set of all integers from $1$ to $n$. The set $T$ is the set of all distinct subsets of $S$, including the empty set $\emptyset$ and $S$ itself. Show that $T$ contains exactly $2^n$ sets.
The sets $A_1, A_2, ..., A_m$, which are not necessarily distinct, are chosen randomly and independently from T, and for each $k$ $(1 \le k \le m)$ the set $A_k$ is equally likely to be any of the sets in $T$.
(i) Write down the value of $P(1 \in A).$
(ii) By considering each integer separately, show that $P(A_1 \cap A_2= \emptyset )= \binom{3}{4}^n$. Find $P(A_1 \cap A_2 \cap A_3 = \emptyset)$ and $P(A_1 \cap A_2 ... \cap A_m = \emptyset)$.
(iii) Find $P(A_1 \subseteq A_2)$, $P(A_1 \subseteq A_2 \subseteq A_3)$ and $P(A_1 \subseteq A_2 \subseteq ... \subseteq A_m)$.
Having successfully worked through all but the last part (and the first probability of the last part), I'm confused at how to extend my method for 2 subsets to more than 2 subsets (although I did find the last 2 probabilities via a different method). I'm asking as I am curious as to how to extend it for more than 2 subsets, and am unable to think of how myself.
So consider all possible pairs $(A_1,A_2)$. $A_1$ and $A_2$ can each take one of $2^n$ values, and so there are $2^{2n}$ such pairs. Now, for $A_1$ to be a subset of $A_2$, it has to be one of the possible subsets of $A_2$ itself. Say $|A_2|=k$ for some $0 \le k \le n$. Then the total number of possibilities for $A_1$ is $$\sum_{k=0}^{n}{{n}\choose{k}}{2^k}=(1+2)^n=3^n$$ by the Binomial theorem. $$\therefore \mathbb{P}(A_1\subseteq{A_2})=\frac{3^n}{2^{2n}}=(\frac{3}{4})^n$$
I can't determine how to extend this approach to m subsets; any insight would be appreciated.