%% LyX 1.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[12pt]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\setlength\parskip{\medskipamount}
\setlength\parindent{0pt}

\makeatletter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
\providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\topmargin}{0.0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\textwidth}{6.5in}
\setlength{\textheight}{9.0in}
\setlength{\parindent}{0in}
\setlength{\parskip}{0.2in}
\setlength{\columnseprule}{0.4pt}

\usepackage{times}
\usepackage{fancyheadings}
\usepackage[dvips]{graphicx}

\makeatother

\begin{document}

\Large

\title{Our First Steps in Probability}

\author{}

\date{}

\maketitle

\pagestyle{fancy} 
\chead{}
\cfoot[]{II.\thepage:  Probability for Kids,
\copyright{2002}, N.  Matloff, University of California, Davis}

Think about the following question:

\begin{quote}
Suppose we deal a 5-card hand from a regular 52-card deck.  Which is
larger, P(1 king) or P(2 hearts)?
\end{quote}

We will be able to solve this later.  For now, we are building the
foundation.

\newpage

\ 

Example:  Roll 2 dice, say a blue one and a yellow one.  

Find P(blue + yellow = 6).

Possible outcomes (blue, then yellow):

\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|c|c|c|c|}
\hline 
1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\
\hline
2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\
\hline
3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\
\hline
4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\
\hline
5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
\hline
6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \\
\hline
\end{tabular}\par}\vspace{0.3cm}

All equally likely!  (Very important.)

Now, which ones are we interested in?


\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|c|c|c|c|}
\hline 
1,1 & 1,2 & 1,3 & 1,4 & {\bf 1,5} & 1,6 \\
\hline
2,1 & 2,2 & 2,3 & {\bf 2,4} & 2,5 & 2,6 \\
\hline
3,1 & 3,2 & {\bf 3,3} & 3,4 & 3,5 & 3,6 \\
\hline
4,1 & {\bf 4,2} & 4,3 & 4,4 & 4,5 & 4,6 \\
\hline
{\bf 5,1} & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
\hline
6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \\
\hline
\end{tabular}\par}\vspace{0.3cm}

5 out of 36 possible outcomes have blue+yellow = 6.  So,

$$
P(blue+yellow=6) = \frac{5}{36} 
$$

\newpage

\

{\bf But what does that really mean?}

Here our ``experiment'' is to roll 2 dice.  

Imagine doing the experiment many, many times, recording the results in
a large notebook:

\begin{itemize}

\item Roll the dice the first time, and write the outcome on the first
line of the notebook.

\item Roll the dice the second time, and write the outcome on the second
line of the notebook.

\item Roll the dice the third time, and write the outcome on the third
line of the notebook.

\item Roll the dice the fourth time, and write the outcome on the fourth
line of the notebook.

\item Etc.  Imagine you keep doing this, thousands of times, filling
thousands of lines in the notebook.

\end{itemize} 

\newpage

\

The notebook might look like this:

\begin{verbatim}

    outcome                blue+yellow = 6?

blue 2, yellow 6                 no
blue 3, yellow 1                 no
blue 1, yellow 1                 no
blue 4, yellow 2                 yes
blue 1, yellow 1                 no
blue 3, yellow 4                 no
blue 5, yellow 1                 yes
blue 3, yellow 6                 no
blue 2, yellow 5                 no
...                              ...
...                              ...
...                              ...

\end{verbatim}

Here 2/9 of the lines say ``yes.''

But after many, many repetitions, approximately 5/36 of the lines will
say ``yes.''

For example, about how may lines will say ``yes'' if we do the
experiment 720 times?  

\newpage

\

This is what probability really is:  In what fraction of the lines does
something of interest happen?

{\bf It sounds simple, but if you always think about this ``lines in the
notebook'' idea, probability problems are a lot easier to solve.}

\newpage

\

So, P(blue+yellow=6) = 5/36.

What about P(blue = 2)?  

You might answer, ``It depends on what the yellow die is.''  But that is
wrong.  You should reason as follows:

\begin{quote}
``If I do this experiment many, many times, approximately what fraction
of the lines will have blue = 2?''
\end{quote}

\begin{verbatim}

    outcome                    blue = 2?

blue 2, yellow 6                 yes
blue 3, yellow 1                 no
blue 1, yellow 1                 no
blue 4, yellow 2                 no
blue 1, yellow 1                 no
blue 3, yellow 4                 no
blue 5, yellow 1                 no
blue 3, yellow 6                 no
blue 2, yellow 5                 yes
...                              ...
...                              ...
...                              ...

\end{verbatim}

\newpage

\

So, what fraction of lines will have say ``yes''?

\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|c|c|c|c|}
\hline 
1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\
\hline
{\bf 2,1} & {\bf 2,2} & {\bf 2,3} & {\bf 2,4} & {\bf 2,5} & {\bf 2,6} \\
\hline
3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\
\hline
4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\
\hline
5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
\hline
6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \\
\hline
\end{tabular}\par}\vspace{0.3cm}

6 of the 36 possibilities have blue = 2, and they are all equally
likely, so about 6/36 = 1/6 of the lines will have blue = 2.

So, P(blue = 2) = 1/6.

Also:

P(blue = 3 and yellow = 6) = 1/36.

P(blue = 3 or yellow = 6) = 11/36.


\newpage

\

Problems to be worked in class:


\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|c|c|c|c|}
\hline 
1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\
\hline
2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\
\hline
3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\
\hline
4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\
\hline
5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
\hline
6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \\
\hline
\end{tabular}\par}\vspace{0.3cm}

P(blue + yellow = 7) = ?

P(blue = 4 and yellow = 1) = ?  

P(blue = 3 or yellow = 3) = ?  

\newpage

\

Let's move to something a bit more complicated:

\begin{quote}

The teacher asks Bill, Ann, Tom, Cathy and Linda to draw straws.  The two
kids who get the two shortest straws get a free ice cream cone.  

P(Cathy get a cone) = ?

P(no boy gets a cone) = ?

P(one boy and one girl get a cone) = ?

\end{quote}

Possible outcomes (be careful!):

\begin{verbatim}

B,A   B,T   B,C   B,L
A,T   A,C   A,L
T,C   T,L
C,L

\end{verbatim}

(We don't include, for example, A,B, since it's the same to us as B,A.
They both get cones.)

There are 10 possible outcomes, and all are equally likely.  So:

P(Cathy get a cone) = 4/10 = 2/5

P(no boy gets a cone) = 3/10

P(one boy and one girl get a cone) = 6/10 = 3/5

\newpage

\

\begin{verbatim}

outcome             Cathy gets a cone?

B,T                       no
T,L                       no
C,L                       yes
A,C                       yes
A,T                       no
...                       ...
...                       ...

\end{verbatim}

After, say, 1,000 repetitions of this experiment, about how many lines
will say ``yes''?  

$\frac{2}{5} \times 1,000 = 200$ lines

\newpage

\

{\bf Homework:}

{\it Main problem:}  Add David to the list of kids who will draw straws for
the two ice cream cones.  (So, there are now 6 kids in all.)  Answer the
same 3 questions as before:  P(Cathy gets a cone) = ?, etc.

{\it Challenge problem:}  Suppose our experiment is to roll 4 dice, not
2.  Say we have one blue die, one yellow, one green, one red.  Find
P(blue+yellow+green+red = 9).  (There are over 1,000 possible outcomes
of this experiment, compared to 36 for the 2-dice example above.  Do NOT
list all 1,000+ outcomes.  Find a shortcut.)

{\bf Next week:}

Conditional probability.


\end{document}

