\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
For each integer between \(1\) to \(n\) inclusive, they are either in a subset of \(S\), an element of \(T\), or not. For each integer there are \(2\) choices, and there are \(n\) integers, this means that \[ \abs *{T} = 2^n, \] as desired.
For each of the integer \(1 \leq t \leq n\), \(t \notin A_1 \cap A_2\) if and only if they cannot be in both of \(A_1\) and \(A_2\), and hence \[ \Prob (t \notin A_1 \cap A_2) = 1 - \left (\frac {1}{2}\right )^2 = \frac {3}{4}, \] and \(A_1 \cap A_2 = \emptyset \) if and only if for all \(1 \leq t \leq n\), that \(t \notin A_1 \cap A_2\). All these events are independent, and hence \[ \Prob (A_1 \cap A_2 = \emptyset ) = \left (\frac {3}{4}\right )^n. \]
By similar reasoning, \[ \Prob (A_1 \cap A_2 \cap A_3 = \emptyset ) = \left (\frac {7}{8}\right )^n, \] and \[ \Prob (A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset ) = \left [1 - \left (\frac {1}{2}\right )^m\right ]^n = \left (1 - \frac {1}{2^m}\right )^n. \]
\(A_1 \subseteq A_2\) if and only if for any \(1 \leq t \leq n\), we have \(t \in A_1 \implies t \in A_2\). For this to happen, either \(t \notin A_1\) (in which case we do not worry about whether \(t\) is in \(A_2\) or not), or \(t \in A_1\) and \(t \in A_2\). This means \[ \Prob (t \in A_1 \implies t \in A_2) = \frac {3}{4}, \] and hence \[ \Prob (A_1 \subseteq A_2) = \left (\frac {3}{4}\right )^n. \]
For any \(1 \leq t \leq n\), \(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m\) means we have \(t \in A_1 \implies t \in A_2 \implies \cdots \implies t \in A_m\). This happens if and only if \(t \in A_i\) gives \(t \in A_j\) for all \(j \geq i\), and this is true if and only if there exists some \(0 \leq k \leq m\), such that for \(1 \leq i \leq k\), \(t \notin A_k\), and for \(k < j \leq m\), \(t \in A_k\).
There are precisely \(m + 1\) choices for such \(k\), and this means \[ \Prob (t \in A_1 \implies t \in A_2 \implies \cdots \implies t \in A_m) = \frac {m + 1}{2^m}, \] and hence \[ \Prob (A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m) = \left (\frac {m + 1}{2^m}\right )^n, \] which gives \[ \Prob (A_1 \subseteq A_2 \subseteq A_3) = \left (\frac {1}{2}\right )^n. \]