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\setcounter{customexample}{-1}
\begin{example}
    Dice: outcomes $1, 2, \dots, 6$.
    \begin{itemize}
        \item $\PP(2) = \frac{1}{6}$
        \item $\PP(\text{multiple of 3}) = \frac{2}{6}
            = \frac{1}{3}$.
        \eqitem \begin{align*}
                \PP(\text{prime or a multiple of
                3})
                &\,\, \cancel{= \frac{1}{3} +
                \frac{1}{2} = \frac{5}{6}} \\
                &= \frac{4}{6} = \frac{2}{3} \\
                &= \frac{1}{3} + \frac{1}{2}
                - \PP(\text{prime \emph{and} a multiple
                of 3}) \\
                &= \frac{1}{3} + \frac{1}{2} -
                \frac{1}{6} = \frac{2}{3}
            \end{align*}
        \item $\PP(\text{not a multiple of 3}) =
            \frac{2}{3}$.
    \end{itemize}
\end{example}

\section{Formal Setup}

\begin{definition*}
    \begin{itemize}
        \item \emph{Sample space} $\Omega$, a set of
            \emph{outcomes}.
        \item $\mathcal{F}$ a collection of subsets of
            $\Omega$ (called \emph{events}).
        \item $\mathcal{F}$ is a
            \emph{$\sigma$-algebra}
            (``sigma-algebra'') if:
            \begin{enumerate}[F1]
                \item $\Omega \in \mathcal{F}$
                \item if $A \in \mathcal{F}$ then
                    $A^c \in \mathcal{F}$ ($A^c :=
                    \Omega \setminus A$)
                \item $\forall$ countable
                    collections $(A_n)_{n \ge 1}$
                    in $\mathcal{F}$ the union
                    \[ \bigcup_{n \ge 1} A_n \in
                    \mathcal{F} \]
                    also.
            \end{enumerate}
    \end{itemize}
\end{definition*}
Given $\sigma$-algebra $\mathcal{F}$ on $\Omega$,
function $\PP : \mathcal{F} \to [0, 1]$ is a
\emph{probability measure} if
\begin{enumerate}[P1]
    \item[P2] $\PP(\Omega) = 1$
    \item[P3] $\forall$ countable collections
        $(A_n)_{n \ge 1}$ of disjoint events in
        $\mathcal{F}$:
        \[ \PP \left( \bigcup_{n \ge 1} A_n \right)
        = \sum_{n \ge 1} \PP(A_n) .\]
\end{enumerate}
(P1 was historically taken to state that $\PP(A)
\ge 0$, but this is already captured by the
notation $\PP : \mathcal{F} \to [0, 1]$).

\bigskip

\noindent
Then $(\Omega, \mathcal{F}, \PP)$ is a
\emph{probability space}.

\subsubsection*{Revisiting dice example}

For a dice we have:
\[ \Omega = \{1, 2, \dots, 6\} \]
\[ \PP(\text{1 or 2 or 3 or 4 or 5 or 6}) = 1 .\]
\[ \mathcal{F} = \mathcal{P}(\Omega) \]

\noindent
\textbf{Question}: Why $\PP : \mathcal{F} \to [0,
1]$ not $\PP : \Omega \to [0, 1]$? \\
\ul{$\Omega$ finite / countable}
\begin{itemize}
    \item In general: $\mathcal{F} = \text{all subsets
        of $\Omega$}$. ($\PP(\Omega)$).
    \item $\PP(2)$ is shorthand for $\PP(\{2\})$.
    \item $\PP$ is determined by
        ($\PP(\{\omega\}), \forall \,\,\omega \in
        \Omega$). (eg unfair dice)
\end{itemize}
\ul{$\Omega$ uncountable}
\begin{itemize}
    \item For example $\Omega = [0, 1]$. Want to
        choose a real number, all equally
        likely.
    \item If $\PP(\{0\}) = \alpha > 0$, then
        \[ \PP \left( \left\{ 0, 1, \frac{1}{2},
        \dots, \frac{1}{n} \right\} \right) = (n +
        1) \alpha \]
        \contradiction if $n$ large as $\PP > 1$.
    \item So $\PP(\{0\}) = 0$, or $\PP(\{0\})$ is
        undefined.
    \item What about $\PP(\{x : x \le
        \frac{1}{3}\})$?
        \begin{itemize}
            \item ? ``Add up'' all $\PP(\{x\})$
                for $x \le \frac{1}{3}$.
        \end{itemize}
\end{itemize}

\begin{example*}
    $\Omega = \{f : \text{continuous on $[0,1] \to
    \RR$, $f(0) = 1$}\}$. What is
    $\PP(\text{differentiable})$?
\end{example*}

\subsection{From the axioms}

\begin{itemize}
    \item $\PP(A^c) = 1 - \PP(A)$.
        \begin{proof}
            $A$, $A^c$ are disjoint. $A \cup A^c =
            \Omega$ and hence
            \[ \PP(A) + \PP(A^c) \stackrel{P3}{=}
            \PP(\Omega) \stackrel{P2}{=} 1 \]
        \end{proof}
    \item $\PP(\emptyset) = 0$.
    \item If $A \subseteq B$ then $\PP(A) \le
        \PP(B)$.
    \item $\PP(A \cup B) = \PP(A) + \PP(B) - \PP(A
        \cap B)$.
\end{itemize}

\subsection{Examples of Probability Spaces}

$\Omega$ finite, $\Omega = \{\omega_1, \dots,
\omega_n\}$, $\mathcal{F} = \text{all subsets}$
uniform choice (equally likely).
\[ \PP : \mathcal{F} \to [0, 1], \quad \PP(A) =
\frac{|A|}{|\Omega|} .\]
In particular:
\[ \PP(\{\omega\}) = \frac{1}{|\Omega|} \,\,
\forall \,\, \omega \in \Omega .\]

\begin{example}
    Choosing without replacement $n$
    indistinguishable marbles labelled $\{1,
    \dots, n\}$. Pick $k \le n$ marbles uniformly
    at random. Here:
    \[ \Omega = \{A \subseteq \{1, \dots, n\} :
    |A| = k\} \qquad |\Omega| = {n \choose k} \]
\end{example}

\begin{example}
    Well-shuffled deck of cards. Uniformly chosen
    \emph{permutation} of 52 cards.
    \[ \Omega = \{\text{all permutation of 52
    cards}\} \qquad |\Omega| = 52! \]
    \[ \PP(\text{first three cards have the same
    suit}) = \frac{52 \times 12 \times 11 \times
    49!}{52!} = \frac{22}{425} \]
    Note: $= \frac{12}{51} \times \frac{11}{50}$.
\end{example}
