\documentclass[12pt,twoside]{article}

\def\handouttype{problemset}

\usepackage{psfig}
\usepackage{changebar}
%\newcommand{\cbstart}{\chgbarbegin}
%\newcommand{\cbend}{\chgbarend}

\input{macros-course}

\def\cent{{\not\!\mathrm{c}}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\askaboutsolutions

\ifshowsolutions 
  \usepackage[dark,first,bottom]{draftcopy}
  \draftcopyName{Solutions}{155}
  \typeout{}
  \typeout{Solutions *INCLUDED*}
  \renewcommand{\solution}[1]{
    \textbf{Solution.} \par %
    #1
    \mbox{} \\
    \mbox{}\hrulefill\hspace*{1ex}\ignorespaces%
    \raisebox{-0.25\baselineskip}{$\square$}%
    \hspace*{1ex}\ignorespaces\hrulefill%
  }
\else
  \typeout{}
  \typeout{Solutions *NOT INCLUDED*}
  \renewcommand{\solution}[1] {
  }
\fi
    

%\renewcommand{\solution}[1]{%
%  \ifshowsolutions %
%%  \cbstart
%%  \cbend
%  \nopagebreak%
%  \else
%%  \vspace*{2in}
%    \fi}

\newcommand{\problembreak}{%
  \ifshowsolutions %
  \newpage
  \else
  \newpage
  \fi%
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\tutref}{tut10}
\newcommand{\tutnum}{10}
\newcommand{\tutdate}{April 14, 2000}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\handout{\tutref}{Mini-Quiz \tutnum}{\tutdate}

\begin{enumerate}
\item Write your name:

\bigskip

\item %
  (Rosen, Section 4.7, Exercise 1) Find the next largest permutation
  in lexicographic order after each of the following permutations.
\begin{enumerate}
\item 1432
\item 12453
\item 6714235
\item 54123
\item 45231
\item 31528764
\end{enumerate}
\solution{
2134, 12534, 6714253, 54132, 45312, 31542678
}

  \vspace*{3in}

\end{enumerate}

\newpage
\thispagestyle{empty}

\mbox{}

\setcounter{page}{0}

\newpage
\handout{\tutref}{Tutorial \tutnum~Problems}{\tutdate}

\begin{problems}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Source: Spring 99

\problem In the O.J. Simpson trial, defense attorney Dershowitz observed
that the fraction of wife batterers that murder their wives is $1/2000$.
Since the probability that a wife batterer would murder his wife is so
low, Dershowitz argued that the fact that Simpson battered his wife,
Nicole Brown Simpson, should be excluded from the evidence.  We will
develop a counter-argument to Dershowitz', by conditioning the probability
on the event that Mrs. Simpson is dead.

We have the following statistical information (all for 1994):
\begin{center}
\begin{tabular}{|l|c|}
\hline
women & 125M\\
married women & 50M\\
wives battered by husband & 3M\\
women murdered & 6K\\
wives murdered by husband & 1.5K\\
\hline
\end{tabular}
\end{center}
The number of battered wives includes those who were murdered by their
husbands. 
% rjr
Also, assume that the act of murder is enough to qualify for battering.
%rjr
Define the events
\begin{eqnarray*}
H & ::= & \mbox{wife is murdered by husband} \\
M & ::= & \mbox{woman is murdered} \\
B & ::= & \mbox{wife is battered by husband}
\end{eqnarray*}

\begin{problemparts}

\problempart Dershowitz' statistical observation can be stated technically as
\[\pr{H \mid B} = 1/2000.\]
Prove it.

\solution{
\[
\pr{H \mid B}
= \frac{\pr{H \intersect B}}{\pr{B}}
=\frac{\pr{H}}{\pr{B}}=\frac{1500/125M}{3M/125M}
=\frac{1500}{3M}=1/2000.
\]}

\problempart Calculate $\pr{H \mid M}$, i.e., the probability that a murdered
woman was murdered by her husband.

\solution{
\[\pr{H \mid M}
= \frac{\pr{H \intersect M}}{\pr{M}}
=\frac{\pr{H}}{\pr{M}}=\frac{1500/125M}{6000/125M}
= \frac{1500}{6000}=1/4.
\]}

\problempart Prove the identity
\[
\pr{C \mid A}  =  \pr{C \intersect D \mid A} + \pr{C - D \mid A}
\]
for arbitrary events $A, C, D$.

\solution{
\begin{eqnarray*}
\pr{C \intersect A} & = & \pr{(C \intersect A) \intersect D}
    + \pr{(C \intersect A) - D},\\
& = & \pr{(C \intersect D) \intersect A}
    + \pr{(C - D) \intersect A}.\\
\end{eqnarray*}
Now dividing both sides of this equation by $\pr{A}$ yields the desired
identity.}

\problempart Calculate $\pr{H \mid B \intersect M}$, i.e., the probability that
a battered, murdered woman was murdered by her husband.  Assume that the
the probability of a wife being murdered by someone \emph{other} than her
husband is not influenced by whether or not she was being battered.

\solution{
Note that
\[
\pr{H \mid B \intersect M}
= \frac{\pr{H \intersect B \intersect M}}{\pr{B \intersect M}}
= \frac{\pr{H}}{\pr{M \mid B} \pr{B}}.
\]

To evaluate $\pr{M \mid B}$, we apply the identity from the previous
problem part:
\[
\pr{M \mid B} = \pr{M \intersect H \mid B} + \pr{M - H \mid B}.
\]
We know that being murdered by someone other than the husband is not
influenced by whether the victim was battered by her husband or not,
namely,
\[\pr{M - H \mid B} = \pr{M - H},\]
and $H \intersect M = H$, so
\[
\pr{M \mid B} =  \pr{H \mid B} + \pr{M - H} =
\frac{1}{2000} + \frac{6K - 1.5 K}{125M}.
\]

Now we have
\begin{eqnarray*}
\pr{H \mid B \intersect M} & = & \frac{1500 / 125M}{(1/2000 +
(6K - 1.5 K)/125M)3M/125M}\\
 & = & \frac{1500}{3M/2000 + 4500 \cdot 3/125}\\
 & = & \frac{1}{1 + 9/125}\ =\ \frac{125}{134}\  \approx\  0.93.
\end{eqnarray*}
}

\problempart How could you counter Dershowitz' argument that evidence
that a husband battered his wife should not be admitted as evidence?

\solution{Being battered increases the probability that a murdered woman
was murdered by her husband from 25\% to 93\%, so the fact of being
battered greatly influences the probability of a husband's guilt.
Assuming it is proper for a jury to consider such probabilities -- this is
questionable by the way -- then Dershowitz' argument that battery is
unlikely to be relevant does not hold up.}

\end{problemparts}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Source fall 97

\problem
Consider playing the {\it Lets Make a Deal}, where there are three prizes:
a ``Gift Certificate'' from Toscanini, a ``Nerd Pocket Protector'', and a 
``Nerd Pride Button''. Your goal is to win the gift certificate. 

\begin{problemparts}
\problempart
Suppose you (the contestant) happen
to know that Carol always reveals the ``Nerd Pocket Protector'', if you 
did not choose the box covering it.

Does this additional information change the odds of winning for you? 
Is there a better strategy? By how much (if any) does this information
increase your chances of winning?

\solution{
The probability of winning the game remains the same, but now there is
more than one optimal strategy.  If Carol reveals the ``Nerd Pride
Button,'' it means that you must have chosen the box covering the
``Nerd Pocket Protector''.  In this case, you win with probability 1
if you switch. If Carole shows you the ``Nerd Pocket Protector'' then
there is $1/2$ probability that you have picked the prize box, so you
win with probability $1/2$ whether you switch or not.  Let NPB be the event
that Carole shows the ``Nerd Pride Button'', and NPP be the event that
Carole shows the ``Nerd Pocket Protector''. Then the 
total probability of winning is:
\begin{eqnarray*}
Pr(win) &=& Pr(win \mid \mbox{NPB})Pr(\mbox{NPB}) + Pr(win \mid\mbox{NPP})
Pr(\mbox{NPP}) \\
        &=& 1 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} \\
        &=& \frac{2}{3}
\end{eqnarray*}
}

\problempart
You are playing according to the original assumptions, but before
you make your initial choice, you happen to see Carol placing a 
``Nerd Pride Button'' underneath Box~1.  What should you do to try to win 
the prize (the ice cream)?  What is the probability that you win?

\solution{
In this case, you should always pick Box 1.  Monty 
will be forced to reveal the location of the other ``Nerd Pride'' pin.
Upon switching, you will be guaranteed to win the ice cream (i.e.
the probability of winning is 1).  If you
pick a box besides 1, Monty can always reveal Box 1 (giving you no
additional information), and you will have a 1/2 chance of winning.
}

\problempart
You are trying one last time, with new rules. This time, you are told that Monty 
will reveal one of the doors you did not choose at random, whether or not it
contains the prize. (In the case that the prize is revealed, you obviously
lose). In this case, what is your optimal strategy and probability of 
winning?

\solution{
It does not matter now whether you switch or stay; you cannot increase your
chances beyond 1/3. A tree diagram shows this nicely. You can also reason
that 1/3 of the time, you have chosen the right door and should stay. 2/3
of the times you will choose the wrong door and be better off switching, but
half of those times you will lose immediately. Thus, they do not contribute to
the 
$\pr{\textrm{\emph{choice was right}} \mid \textrm{\emph{did not lose yet}}}$
that we are really considering.
}

\end{problemparts}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Source fall 94

\problem 
Consider the {\it Lets Make a Deal} game with the following assumptions
on how host Monty Hall hides the prize and on how you initially choose a door.
In each case, you follow the ``switch strategy'', and in each case Monty
reveals a loser after your initial choice.  State whatever assumptions
you need about how Monty reveals the loser.
For each case, determine the probability that you will win the prize by 
listing all events in the sample space 
along with the probabilities that they occur.
\begin{problemparts}
\problempart
Monty Hall 
chooses a door to contain the prize with some non-uniform distribution --
he hides the prize behind door number $i$ with some probability $p_i$, 
for $i=1,2,3$ and $\sum_i p_i =1$.
You, the contestant, (not knowing the $p_i$'s) follow the 
strategy of selecting a door
uniformly at random -- you choose door $i$ with probability $1/3$.
\problempart
Monty Hall chooses a door behind which to 
hide the prize according to the uniform distribution 
but you the contestant select a door according to some non-uniform
distribution -- you select door $i$ with some probability $p_i$, 
$\sum_{i=1}^3 p_i =1$.
\end{problemparts}

\solution{ Since you always follow the ``switch'' strategy, the
probability of winning the prize is the same as the probability that
you did not choose the winning door as your first choice.  In both
parts, we can model this simplified problem with a sample space of
nine points labeled as $(w_i,c_k)$, with $i,k\in \{1,2,3\}$, and where
$(w_i,c_k)$ corresponds to the event that the prize is behind door $i$
and your first choice is door $k$.  The winning probability is then
$1-(w_1,c_1)-(w_2,c_2)-(w_3,c_3)$
\begin{enumerate}
\item In this part, we have $P(w_i,c_k) = \frac{1}{3}p_i$.  So the
probability of winning is $1-\frac{1}{3}=\frac{2}{3}$.
\item In this part, we have $P(w_i,c_k) = \frac{1}{3}p_i$.  So the
probability of winning is also $\frac{2}{3}$.
\end{enumerate}

}

\end{problems}

\end{document}

\end{input}
















