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\begin{document}

\noindent{\heada Chapter 6 - Probability}

\bigskip

\noindent In Chapters 1 and 2 we talked how to collect data and how
to conduct an experiment.  In Chapters 3-5 we covered how to
describe the data which you have collected using numerical and
graphical summaries.  In the rest if the course, we turn our
attention to {\bf Inferential Statistics}, the process of taking
results about a sample and generalizing them to the population from
which the sample was taken.

\bigskip

\noindent We already know that we can infer things about a
population from {\bf random samples}.

\bigskip

\noindent In order to fully understand inferential statistics, we
need the language of probability, which is the topic of this
chapter.

\bigskip

\noindent {\bf \underline{Terminology}:}
\begin{enumerate}
\item {\bf Outcome} - one possible value of a variable

\item {\bf Sample Space} - the set of all possible outcomes of a variable.

\item {\bf Event} - a group of one or more outcomes; An event `occurs' when one of the
outcomes in the event occurs.
\end{enumerate}
\vspace{0.1in}

\underline{EXAMPLES}: \vspace{-0.05in}
\begin{itemize}
\item Flip a coin once and measure the face of the coin.  The sample space of all possible
outcomes is
    $$S=\{\underline{\hspace{2in}}\}.$$

The event that a head occurs is
    $$A=\{\underline{\hspace{2in}}\}.$$


\item Consider drawing a single card from a deck of 52 playing
cards and recording the rank and suit of the card. The sample space
of all 52 possible outcomes is
    $$S=\{\underline{\hspace{2in}}\}.$$
The event that the card is a heart is
    $$A=\{\underline{\hspace{2in}}\}.$$

\item Flip a coin twice.  The sample space of all possible
outcomes is
    $$S=\{\underline{\hspace{2in}}\}.$$
The event that at least one head occurs is
    $$A=\{\underline{\hspace{2in}}\}.$$


\item Count the number of buffalo observed at a certain location
in YNP.   The sample space of all possible outcomes is
    $$S=\{\underline{\hspace{2in}}\}.$$

The event that more than ten buffalo are observed is
    $$A=\{\underline{\hspace{2in}}\}.$$

\item Measure the time it takes for some randomly chosen human
being to run a mile.   The sample space of all outcomes is

    $$S=[\underline{\hspace{2in}}).$$

Consider the event that the mile is completed under 5 minutes,
    $$S=[\underline{\hspace{2in}}).$$

\end{itemize}

\bigskip

\noindent {\bf \underline{BASICS OF PROBABILITY}:}

\noindent {\bf Probability} is a ``long-term" relative frequency
or proportion.  The probability of an event $A$, written $P(A)$,
is the proportion of times an outcome in the event occurs in many
identically-repeated trials.

\begin{verse}
\noindent {\bf \underline{QUESTIONS}:}
\begin{enumerate}
\item Toss a fair coin. How do you know $P$(Head)=0.5?
\vspace{0.05in}


\hspace{0.55in}
\begin{tabular}{r|cccccc}
Toss & $1^{\rm st}$ & $2^{\rm nd}$ & $3^{\rm rd}$ & $4^{\rm th}$ &
$5^{\rm th}$ & etc.\\ \hline
Proportion of heads & & & & & & \\
\end{tabular}

\vspace{0.2in}

\item Net worth is defined as the current value of one's assets less liabilities.
If the threshold level of being wealthy is having a net worth of \$1
million or more, then 3.5 million of the 100 million households in
America are considered wealthy (from ``The Millionaire Next Door:
The Surprising Secrets of American's Wealthy", {\em The New York
Times on the Web}, 1996).  Randomly draw a household from the US.
How do we know that $P($getting millionaire$)=.035$?


\vspace{0.5in}


\end{enumerate}

\end{verse}


\noindent {\bf The Axioms of Probability}
\begin{enumerate}
\item {\fbox {\bf 0 $\le P(A) \le$ 1}}


\noindent How often an event $A$ occurs must be somewhere between
``never" (probability = 0) and ``always" (probability =1).






\item {\bf Addition Rule for Disjoint Events} \ \ $\longrightarrow$ \ \ {\fbox {\bf P(A \underline{or} B) = P(A) + P(B)}}

\begin{itemize}
\item If the events $A$ and $B$ are {\it disjoint}, then $P(A$ or $B) = P(A) + P(B)$.
\item Two events are {\bf disjoint} (or {\bf mutually exclusive}) if they cannot possibly occur simultaneously.
\end{itemize}



\begin{verse}
\noindent {\bf \underline{EXAMPLE}:} Suppose we are drawing a
single card from a deck.  Let $A=\{2\heartsuit,...,$Ace
$\heartsuit\}$ and $B=\{2\diamondsuit,...,$Ace $\diamondsuit\}$
Then $A$ and $B$ are disjoint and the probability of getting a
heart or a diamond on a draw from a deck of 52 cards is

$$P(A)=\underline{\hspace{2in}}$$
$$P(B)=\underline{\hspace{2in}}$$
$$P(A \mbox{~~or~~} B)=\underline{\hspace{2in}}$$.
\end{verse}

\vspace{0.1in}

\item {\bf Complement Rule} \ \ $\longrightarrow$ \ \ {\fbox {\bf P(\underline{not} A) = 1 - P(A)}}

\begin{itemize}
\item The language ``not $A$" is the same as the ``complement of $A$".
\item Either $A$ occurred or $A$ did not occur and they are disjoint, so their probabilities
must sum to 1.  $P(A) + P($\underline{not} $A) = 1$.
\end{itemize}

\begin{verse}
\noindent {\bf \underline{EXAMPLE}:} Suppose we are drawing a single
card from a deck.  Let $A=\{2\heartsuit,...,$Ace $\heartsuit\}$.
Then (not A)=$\{2 \clubsuit, 3 \clubsuit,$ ... ,Ace $\clubsuit, 2
\diamondsuit, 3 \diamondsuit$, ... , Ace $\diamondsuit, 2
\spadesuit, 3 \spadesuit,$ ... , Ace $\spadesuit\}$.


$$P(A)=\underline{\hspace{2in}}$$
$$P(\mbox{not~~}A)=\underline{\hspace{2in}}$$

\end{verse}


\vspace{0.1in}

\item {\bf Multiplication Rule} \ \ $\longrightarrow$ \
\ {\fbox {\bf P(A \underline{and} B) = P(A$|$B) $\times$ P(B)}}

\begin{itemize}
\item $P(A|B)$ is the probability of A occurring \underline{given}
you know B has occurred. \vspace{0.1in}



\noindent {\bf \underline{EXAMPLES}:}
\begin{enumerate}

\item Suppose we are drawing a single card from a deck.  Let $A=\{Q
\spadesuit\}$ and $B=\{2\spadesuit,...,$Ace $\spadesuit\}$.
$$P(A)=\underline{\hspace{2in}}$$
$$P(B)=\underline{\hspace{2in}}$$
$$P(A|B)=\underline{\hspace{2in}}$$
$$P(B|A)=\underline{\hspace{2in}}$$
$$P(A \mbox{~~and~~}B)=\underline{\hspace{2in}}$$
$$P(B \mbox{~~and~~}A)=\underline{\hspace{2in}}$$

\item Suppose that there are 5 wealthy households (having a net worth of over a million dollars) out of the 21 households in some rural county in Montana.
From the tax rolls, randomly choose two households from this county.
Let $B$ be the event that the first household is wealthy. Let $A$ be
the event that the second household is wealthy.
$$P(A)=\underline{\hspace{2in}}$$
$$P(B)=\underline{\hspace{2in}}$$
$$P(A|B)=\underline{\hspace{2in}}$$
$$P(A \mbox{~~and~~}B)=\underline{\hspace{2in}}$$

\item Is randomly choosing two households a SRS of households from
this rural Montanan county?

\bigskip

\item On gradation day at a large university, one graduate is selected at random. Let
$A$ represent the event that the student is an engineering major,
and let $B$ represent the event that the student took a calculus
course in college. Which probability is greater, $P(A|B)$ or
$P(B|A)$?

\end{enumerate}


\newpage
\item The events A and B are {\bf independent} if the knowledge
that one of the events occurred (or did not occur) does not change
the probability of the other event occurring (or not occurring),
$$P(A|B) = P(A)$$




\item {\bf Multiplication Rule for Independent Events} \ \ $\longrightarrow$ \ \ {\fbox {\bf $P(A$ \underline{and} $B) = P(A) \times P(B)$}}
if $A$ and $B$ are {\bf independent}.

\bigskip


{\bf \underline{EXAMPLES}:}
\begin{enumerate}
\item Toss a coin twice.
\begin{enumerate}
\item Let $B$ be the event of a head on the first toss.   Let $A$ be
the event of a head on the second toss.
$$P(A)=\underline{\hspace{2in}}$$
$$P(B)=\underline{\hspace{2in}}$$
$$P(A|B)=\underline{\hspace{2in}}$$
$$P(A \mbox{~~and~~}B)=\underline{\hspace{2in}}$$

\item Are the coin tosses independent?
\vspace{0.05in}
\end{enumerate}

\item Suppose that there are 3.5 million millionaire households
out of the 100 million households in the US. From the rax rolls,
randomly choose 2 households from the US. Let $B$ be the event that
the first household is a millionaire. Let $A$ be the event that the
second household is a millionaire.

\begin{enumerate}
\item
$$P(A)=\underline{\hspace{2in}}$$
$$P(B)=\underline{\hspace{2in}}$$
$$P(A|B)=\underline{\hspace{2in}}$$
$$P(A \mbox{~~and~~}B)=\underline{\hspace{2in}}$$

\item Are $A$ and $B$ independent?

\bigskip

\item When choosing two households from the total of 21 households in some Montanan county,
are $A$ and $B$ independent?

\bigskip
\end{enumerate}


\end{enumerate}


\end{itemize}
\end{enumerate}



\noindent {\bf \underline{An Example using all of the rules of
probability}:}

\noindent Suppose that the sample space of human blood types is
$S=\{O, A, B, AB\}$.   For the American population, the following
table gives the probabilities for each:

\bigskip

{\bf American blood type distribution:}\\
\begin{tabular}{l|cccc} \hline
Blood type & O & A & B & AB \\ \hline
U.S. probability & 0.45 & 0.40 & 0.11 & ? \\ \hline
\end{tabular}
\vspace{0.1in}

\bigskip

\noindent Assume that the following table gives the human blood type
probabilities for the population in China:

\bigskip

{\bf Chinese blood type distribution:}\\
\begin{tabular}{l|cccc} \hline
Blood type & O & A & B & AB \\ \hline China probability & 0.35 &
0.27 & 0.26 & 0.12 \\ \hline
\end{tabular}
\vspace{0.1in}

\begin{enumerate}
\item In the United States, what is the P(AB)? \vspace{0.3in}


\item Maria has type B blood. She can safely receive blood
transfusions from people with blood types O and B. What is the
probability that a randomly chosen American can donate blood to
Maria? \vspace{0.6in}



\item Bozeman Deaconess Hospital needs a donor with type A blood.
Ten American donors come in that day.  What is the probability
that the first nine people do not have type A blood but that the
10th person does have type A blood.

\vspace{1in}

\item What is the probability that at least one of the ten people
has type A blood?

\vspace{1in}

\item Choose an American and a Chinese at random, independently of each other. What is the
probability that both have type O blood? \vspace{0.75in}

\item What is the probability that both have the same blood type?
\vspace{1in}



\end{enumerate}




\section*{Exercises} 6.1 on p258: 1,3, 5, 9

\noindent 6.2 on p263: 15, 17

\noindent 6.3 on p273: 19, 21

\section*{Reading} Sections 6.1-6.3

\end{document}