

\begin{slide}
\subsection{\bf Immediate Deductions from}
\subsection{\bf Axioms of Probability}
\begin{itemize}
\item Note that it follows from the axioms that
\item (a) $0 \le P(S) \le 1$: any $S$

\begin{description}
\item{\bf Proof}:\ \ $P(S) \ge 0$ was postulated.  Take $S_1 = S$, 
$S_2 = Y-S=S^*$ (the complement of $S$) in Axiom 3.

$P\left(S_1\right) + P\left(S_2\right) = P(Y) = 1$,\ \ but

$P\left(S_2\right) \ge 0 \Rightarrow P\left(S_1\right) = P(S) \le 1$.
\end{description}

\item (b) $P\left(S+1 + S_2 \right) \le P\left(S_1\right) + P(S_2)$: 
any $S_1$, $S_2$ with equality if $S_1$ and $S_2$ are disjoint.

{\bf Proof:}
\begin{eqnarray*}
S_1 + S_2 & = & \left( S_1 - S_1\,S_2 \right) + S_2 \\
P\left( S_1 + S_2 \right) & = & P\left( S_1 - S_1\,S_2 \right) + 
P\left( S_2 \right) 
\end{eqnarray*}
are sets disjoint.  But $P\left(S_1 - S_1\,S_2\right) \le P\left(S_1\right)$
--- see (a).  This proves desired result.
\end{itemize}
\end{slide}

