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\title{Stat 311-Lecture 01, Probability}
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%\author{\Large Probability}
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Sept 13, 2005


\begin{enumerate}
\item {\em Definition.}


An {\em experiment} is any action whose outcome is subject to
uncertainty (randomness).

The {\em sample space} $S$ of an experiment is the set of all
possible outcomes of the experiment.

An {\em event} $A$ is a subset of the sample space, i.e. $A \subset
S$.


Two events $A$ and $B$ are {\em mutually exclusive} (disjoint) if
they have no outcomes in common, i.e. $A \cap B = \phi$.



\item Given a sample space $S$,
probability $P$ is a function that satisfies three axioms:

(i) $P(A) \geq 0$,

(ii) P(S) = 1,

(iii) For disjoint events $A_1$ and $A_2$, $P(A_1 \cup A_2) = P(A_1)
+P(A_2)$.



\item {\em Problem.}


You flip a biased coin repeatedly until you obtain a head. The probability of
obtaining a head is $p > 0$ at each toss. Then we can show that
$P(S) =1$.



\item Using the axioms of probability, we can show the following theorems



(i) If $A \subset B$, $P(A) \leq P(B)$.

(ii) $P(A) = 1 - P(A^c)$.

(iii) If $A$ and $B$ are mutually exclusive, then $$P(A \cap B)=0.$$

(iv) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.



\item {\em Problem.}

Prove $$P(A \cap B) \geq P(A) + P(B) - 1,$$
when $A,B$ are two events in the same sample space.\\

{\it proof.} We have shown in the class that
$$P(A \cup B) = P(A)+P(B)-P(A \cap B).$$ But $P(A \cup B) \leq 1$. So by
rearranging the terms, $P(A \cap B) \geq P(A) + P(B) -1$. This is
called the {\it Bonferroni inequality}.

\item {\em Problem.}

Show $P(A_1 \cup A_2 \cup A_3) \leq P(A_1) + P(A_2) + P(A_3)$.\\

{\it proof.} $P(A_1 \cup B) = P(A_1) + P(B) - P(A_1 \cap B) \leq
P(A_1) + P(B)$. Now let $B=A_2 \cup A_3$. Then similarly $P(B) \leq
P(A_2) + P(A_3)$. So $P(A_1 \cup A_2 \cup A_3) \leq P(A_1) + P(B)
\leq P(A_1) + P(A_2) + P(A_3)$.

\item {\em Problem.}

For events $A$ and $B$, find formulas for the probabilities of the
following events in terms of the quantities $P(A),P(B)$
and $P(A \cap B)$ only.\\
(i) either $A$ or $B$ but not both \\
(ii) at most one of $A$ or $B$

{\it solution.} (i) $P((A \cup B)\backslash(A \cap B))=P(A \backslash B) + P(A \backslash B) = P(A)+P(B)-2P(A\cap B).$ \\
(ii) Let $D$ be the event of {\it at most one of $A$ or $B$}. The
complement of $D$ is the event of {\it both $A$ and $B$}, i.e. $D^c=
A \cap B$. So $P(D)= 1-P(D^c)= 1-P(A \cap B).$


\item Two players $A$ and $B$ flip a biased coin alternately and the first player
to obtain a head wins. The probability of obtaining a head is $p >
0$ at each toss. If $A$ flips first, find the probability that $A$
wins the game.

{\it Solution.} The event of A wins the game can be decomposed into
$$A \mbox{ win } = \cup_{i=1}^{\infty} \{ \mbox{ A wins on the
}i\mbox{-th toss}\}.$$ Then we have
\bq P(A \mbox{ wins }) &=& p + (1-p)^2p + (1-p)^4p + \dots\\
&=& \frac{p}{1-(1-p)^2}.\eq

\end{enumerate}



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