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\begin{document}

\recn{10}{11/13}{November 13, 1998}


\section{A Dice(y) Problem}

This section should be a review of material covered in Tuesday's
lecture.

There is a game involving three dice and two players.  The dice are
not normal; rather, they are numbered as shown in
Figure~\ref{fig:dice}.  Each hidden face has the same number as the
opposite, exposed face.  As a result, each die has only three distinct
numbers, and each number comes up $\frac{1}{3}$ of the time.

\begin{figure}[htbp]
\centerline{\psfig{figure=dice.eps,height=1.5in}}
\caption{\em This figure shows the strange numbering of the three dice
in the Dice Problem.  The number on each concealed face is the same as
the number on the exposed, opposite face.}
\label{fig:dice}
\end{figure}

In the game, the first player can choose any one of the three dice.  Then
the second player chooses one of the two remaining dice.  They both
roll and the player with the higher number wins.  Which of the three
dice should player one choose?  That is, which of the three dice is
best?

For example, die B is attractive, because it has a 9, the highest
number overall; on the other hand, it also has a 1, the lowest number.
Intuition gives no clear answer!  We can solve the problem with the
same four-step method used in the Monty Hall problem.

\begin{claim}
Die A beats die B more than half of the time.
\end{claim}

\begin{proof}
The claim concerns the experiment of throwing dice A and B.

{\em Step 1: Find the Sample Space. } The sample space for this
experiment is indicated by the tree diagram in
Figure~\ref{fig:diceAB}.

{\em Step 2: Define Events of Interest. } We are interested in the
event that die A comes up greater than die B.  The outcomes in
this event are marked ``A'' in the figure.

{\em Step 3: Compute Outcome Probabilities. } To find outcome
probabilities, we first assign probabilities to edges in the tree
diagram.  Each number comes up with probability $\frac{1}{3}$,
regardless of the value of the other die.  Therefore, we assign all
edges probability $\frac{1}{3}$.  The probability of an outcome is the
product of probabilities on the corresponding root-to-leaf path; this
means that every outcome has probability $\frac{1}{9}$.

{\em Step 4: Compute Event Probabilities. } The probability of an
event is the sum of the probabilities of the outcomes in the event.
Therefore, the probability that die A comes up greater than die
B is $\frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9} +
\frac{1}{9} = \frac{5}{9}$.

As claimed, the probability that die A beats die B is greater than
half.

\begin{figure}[htbp]
\centerline{\psfig{figure=diceAB.eps,height=2.0in}}
\caption{\em This is the tree diagram arising when die A is played
against die B.  Die A beats die B with probability $\frac{5}{9}$.}
\label{fig:diceAB}
\end{figure}
\end{proof}


\begin{figure}[htbp]
\centerline{\psfig{figure=/diceBC.eps,height=2.0in}}
\caption{\em Die B beats die C with probability $\frac{5}{9}$.}
\label{fig:diceBC}
\end{figure}


\begin{figure}[htbp]
\centerline{\psfig{figure=diceCA.eps,height=2.0in}}
\caption{\em Die C beats die A with probability $\frac{5}{9}$.  Amazing!}
\label{fig:diceCA}
\end{figure}

\begin{claim}
Die B beats die C more than half of the time.
\end{claim}

\begin{proof}
See the tree diagram in Figure~\ref{fig:diceBC}; the proof is the same
as for the preceding claim.

\end{proof}

We have shown that A beats B and that B beats C.  From these
results, we might conclude that A is the best die, B is second
best, and C is worst.  But this is totally wrong!

\begin{claim}
Die C beats die A more than half of the time!
\end{claim}

\begin{proof}
See the tree diagram in Figure~\ref{fig:diceCA}.
\end{proof}

Die A beats B, B beats C, and C beats A!  Clearly,
there is no ``transitive law'' here!  This means that no matter what
die the first player chooses, the second player can choose a die that
beats it with probability $\frac{5}{9}$.  The player who picks first is
always at a disadvantage!


\section{Some Identities and Inequalities}

\subsection{The results}

In lecture we showed
\[\pr{A \union B} = \pr{A} + \pr{B} \mbox{ for disjoint } A,B.\]
This first fact generalizes to any countable union:
\[\pr{\Union_n A_n} =  \sum_n\pr{A_n} \mbox{ for pairwise disjoint } A_n.\]

Some other useful identities and inequalities are:
\begin{eqnarray}
\pr{A-B} & = & \pr{A} - \pr{A \intersect B}, \mbox{ so} \label{a-b}\\
\pr{A \intersect B} & \le & \pr{A}, \label{ab<a}\\
\pr{\overline{B}} & = & 1 - \pr{B}, \label{1-b}\\
\pr{A \union B} & = & \pr{A} + \pr{B} - \pr{A \intersect B}, \mbox{ so}\\
\pr{A \union B} & \le & \pr{A} + \pr{B}.
\end{eqnarray}

You can go over the proofs of some of these.

For example,
the last fact is known as \emph{Boole's Inequality}, where equality holds
if and only if $Pr(A \intersect B) = 0$.  Boole's Inequality can also be
extended to any countable number of events:

\begin{theorem}[Boole's Inequality]\label{bool}
$\pr{\Union_n A_n} \leq \sum_n \pr{A_n}$.
\label{thm:n-union}
\end{theorem}
\begin{proof}
Note that
\[\sum_{p \in \Union_n A_n}\pr{p} \le \sum_n\sum_{p \in A_n}\pr{p}\]
because every term in the sum on the lefthand side occurs on the righthand
side.  But the lefthand side is $\pr{\Union_n A_n}$ by definition, and the
righthand side is $\sum_n \pr{A_n}$.
\end{proof}

\begin{corollary}
\label{cor}
$\pr{\Intersect_n A_n} \ge 1 - \sum_n \pr{\overline{A_n}}$.
\begin{proof}

\[\begin{array}{rcll}
\pr{\Intersect_n A_n} & = & 1 - \pr{\overline{\Intersect_n A_n}}
            & \mbox{by equation~(\ref{1-b})}\\
&=& 1 - \pr{\Union_n \overline{A_n}} & \mbox{by DeMorgan's Law}\\
&\ge & 1 - \sum_n \pr{\overline{A_n}} & \mbox{by Boole's Inequality}.
\end{array}\]
\end{proof}

\end{corollary}


\subsection{Example: Wiring circuits}

Suppose you are wiring up a circuit and need to make a total of $n$
connections.
Suppose that we can assume that any particular connection is made {\em
incorrectly} with probability $p$, for some $0 \leq p \leq 1$.  That
is, for $1 \le i
\le n$,
\[\pr{\mbox{$i^{th}$ connection is wrong}} = p.\]
What is the probability that the circuit is wired correctly, i.e., there
are no incorrect connections?
 
Let $A_i$ denote the event that connection $i$ is
made \emph{correctly}.  Then $\overline{A_i}$ is the event that connection
$i$ is made incorrectly, so $\pr{\overline{A_i}} = p$.  Now, using
Corollary \ref{cor}:

\[\pr{\mbox{all connections are OK}} = \pr{\Intersect_{i=1}^n A_i} \ge 1 -
\sum_{i=1}^n \pr{\overline{A_i}} = 1-np.\]

What about an upper bound?  By equation~(\ref{ab<a}),
\[\pr{\mbox{all connections are OK}} = \pr{\Intersect_{i=1}^n A_i} \le
\pr{A_1} = 1-p.\]

So for example, if $n = 10$ and $p = 0.01$, we get the following
bounds:
\[0.9 = 1 - 10\cdot 0.01 \le \pr{\mbox{all connections are OK}} = \le 1-0.01 = 0.99.\]

So we have concluded that the chance that all connections are okay is
somwhere between 90\% and 99\%.  Although we don't have enough information
to calculate the probability of individual sample points, we were able to
conclude something useful by using simple facts from set theory.

Could it actually be as high as 99\% ?  Yes, if the errors occur in such a
way that whenever you make one connection wrong, all the connections are
wrong.  Formally, the sample space could consist of two atomic events, one
corresponding to ``all OK'' with probability $99/100$ and one with
corresponding to ``none OK'' with probability $1/100$.  Then $A_1
= A_2 = \ldots = A_{10} = \Intersect_{i=1}^{10}A_i$, so
$\pr{\mbox{all OK}} = \pr{A_i} = 0.99$.

\textbf{Class Exercise}:  A more natural sample space might be
\[
    S = \set{0,1}^{10}
\]
where each point $s = (b_1,b_2,\ldots,b_{10})$ describes the bad set of
connections in a circuit: $b_i= 0$ if the $i$th connection is bad and
$b_i = 1$ if it is OK.  Now what is $A_1$?  Is it equal to $A_2$?
How might we assign probabilities to the points in
$S$ so $\pr{\mbox{all OK}} = \pr{A_1} = 0.99$?

\textbf{Solution}:
$A_i = \set{(b_1,\dots,b_{10}) \in S : b_i = 1}$, so $A_1\neq A_2$.
$\pr{1^{10}}=0.99,\ \pr{0^{10}}=0.01$, and probability of all other points
is zero.

\medskip

Could it be 90\% ?  Yes, suppose the errors are such that we never make
two wrong connections.  So the events $\overline{A_i}$ are all disjoint and the
probability of getting it right is
\[
\pr{\Intersect A_i}  =  1 - \pr{\Union \overline{A_i}}
     = 1 - \sum_{i=1}^{10} \pr{\overline{A_i}} = 1-10\cdot 0.01 = 0.9.
\]

\textbf{Class Exercise}: How might we assign probabilities to the points
in $S$ above so $\pr{\mbox{all OK}} = 0.9$ but $\pr{A_i} = 0.99$?

\textbf{Solution}: $\pr{s}=0.01$ if $s$ contains 9 ones and 1 zero,
$\pr{1^{10}}=0.9$, and probability of all other points is zero.

\medskip

\subsection{Example: Nontransitive dice again}

The three dice A, B, C have the property that on any roll of dice A and B,
the probability that A beats B is 5/9, on a roll of B and C the
probability that B beats C is 5/9, and likewise on a roll of C and A, the
probability that C beats A is 5/9.  How high can we make this probability?
That is, can we construct dice so that the probability of A beating B, B
beating C, C beating A are all arbitrarily high? --- say 3/4 each?

To answer this, let's roll all three dice.  Define $A$ to be the event
that A beats B, $B$ the event that B beats C and $C$ the event that C
beats A.
\begin{theorem}
\[\min \set{\pr{A}, \pr{B}, \pr{C}} \leq \frac{2}{3}.\]
\end{theorem}

\begin{proof}
\[\begin{array}{rcll}
\pr{A \intersect B \intersect C}
    & = & 1 - \pr{\overline{A} \union \overline{B} \union \overline{C}} 
          & \mbox{by equation~\ref{1-b} and DeMorgan}\\
    & \geq & 1 - (\pr{\overline{A}}+ \pr{\overline{B}} + \pr{\overline{C}}
                & \mbox{by Boole's Inequality}\\
    & = & \pr{A} + \pr{B} + \pr{C}) - 2 & \mbox{by equation~\ref{1-b}}.
\end{array}\]
But $\pr{A \intersect B \intersect C} = 0$,
since it is impossible that on any single roll, die A is higher than die B,
\emph{and} B is higher than C, {\em and} C is higher than A.
Hence
\[2  \geq  \pr{A} + \pr{B} + \pr{C} \geq 3 \min \set{\pr{A},\pr{B},\pr{C}}\]
and therefore
\[2/3 \geq \min \set{\pr{A}, \pr{B}, \pr{C}}.\]
\end{proof}

\textbf{Extra Credit Problem}: Show that the 2/3 bound can be improved to $p
= (\sqrt{5} - 1)/2 = 0.618\ldots$, and that this is the minimum bound.

\section{Conditional probability and Bayes rule}

Recall the formula for conditional
probability:
\[
    \pr{A \mid B} = \frac{\pr{A \intersect B}}{\pr{B}}
\]

As an example, consider rolling two dice. 
What is the probability of having rolled at
least one 3, given that total of the two dice is 8? 

Assume each die gives each outcome (i.e., 1,2,3,4,5, or 6) with probability
1/6, no matter what the outcome of the other die is.
Let $A$ be the event that
total seen is 8, and let $B$ be the event that at least one 3 has
been rolled. The sample space is $S=\set{(i,j) : i,j \in
\set{1,2,3,4,5,6}\, }$, so that $\card{S}=36$.
\[
A = \set{(2,6),(3,5),(4,4),(5,3),(6,2)}, \quad
B\intersect A = \set{(3,5),(5,3)},
\]
Each outcome has probability 1/36,
so $\pr{A} = 5/36$ and $\pr{B\intersect A} = 2/36$. Therefore $\pr{B|A} =
\frac{2/36}{5/36} = 2/5$.

Are $\pr{A|B}$ and $\pr{B|A}$ related? Yes, both relate to $\pr{A\intersect B}$:
\[
\pr{A|B} = \frac{\pr{A\intersect B}}{\pr{B}}, \mbox{ and }
\pr{B|A} = \frac{\pr{B\intersect A}}{\pr{A}},
\]
so
\[
\pr{A|B}\pr{B} = \pr{A\intersect B} = \pr{B|A}\pr{A}.
\]
This yields the famous {\em Bayes Rule:}
\[
\pr{A|B} = \frac{\pr{B|A}}{\pr{B}} \pr{A}.
\]

{\em Example: } At a school 40\% of students are male, and 60\% of males
and 80\% of all students play sports. What is the probability that a
randomly chosen student is a male, given that the student plays sports?

Let $A$ be the event that a student is male.  Let $B$ be the event that a
student plays sports.  We are given $P(A) = 40\%$, $P(B)=80\%$ and
$\pr{B|A}=60\%$. We want to compute $\pr{A|B}$. By Bayes Rule
\[
\pr{A|B} = \frac{\pr{B|A}}{P(B)} \pr{A} = \frac{0.6\times 0.4}{0.8} = 30\%{.}
\]

\section{Simple Dice Games}

Here are some simple problems based on tossing ordinary 6-sided dice.
Consider the experiment of throwing two dice.
This has a representation as a 2-level tree, where each level
represents one of the two dice.
Draw the tree, and give outcome probabilities. 
Now consider probabilities for various events, and work them out using
the methods from lecture, and the equations given earlier today.

\begin{enumerate}
\item
What is the probability that at least one is a $4$?

A: ${11 \over 36}$

\item
If the first one is a $4$, then
what is the probability that the second one is also a $4$?

A: ${1 \over 6}$.

\item
If at least one of them is a $4$, then 
what is the probability that the other one is also a $4$?

A: $\pr{ \mbox{ two $4$s } \mid \mbox{ at least one is } $4$} = {1 \over 11}$.

\item
What is the probability that both are $\leq 4$?

A: ${4 \over 9}$

\item
What is the probability that the max is exactly $4$?

A: ${7 \over 36}$
\end{enumerate}

You can also invent others.

\section{Monty Hall and State Machines}

Draw a new tree for the Monty Hall problem, this time with four
stages.
This time stage 4 is different from what was considered in Tuesday's
lecture.
\begin{enumerate}
\item
Host chooses where prize goes, probability ${1 \over 3}$ for each
door.
\item
Contestant chooses a door, probability ${1 \over 3}$ for each door.
\item
Host picks a door to open, with equal probability for each possible door.
\item
Contestant chooses whether to switch, with probability ${1 \over 2}$
for ``yes'' or ``no''.
\end{enumerate}

It may be helpful to regard this diagram as representing a state
machine, where the states are the nodes of the tree and the
transitions go from each node to its children.
However, now the transitions have associated weights, assigned in such
a way that for each non-leaf node, the sum of the weights is 1.
So this is a kind of ``probabilistic state machine''.
This is just a way to view this experiment -- we're not defining
invariants or doing any of the other things that we ordinarily do with
state machines.

{\bf Question 1:}
In this experiment, what is the probability that the contestant wins?

A: ${1 \over 2}$.
Derive this in the usual way.
You can then observe that whatever happens at stages 1-3 doesn't
matter at all, if the contestant chooses randomly at the end.
In each situation in which he arrives, he will still have a 
${1 \over 2}$ chance of success.

The tree you just considered has probabilities attached to all the
edges.
In lecture, we considered almost the same tree, but without
probabilities on the stage 4 edges.
Instead, the probabilities were replaced by specified choices (always
switch, or always stick).
Or, you can think of this as two different probability distributions
-- each with probability $1$ for one of the possibilities and
probability $0$ for the other.

We can consider what happens when we replace other probabilistic
choices by specified choices.
What effects do various such specified choices have on the probability
of winning?

{\bf Question 2:}
Fix stage 1 at ``door 1'' and stage 4 at ``switch''.

A: Now the prize is always behind door $1$, which sounds like it should
give a big advantage to the contestant.
However, if the contestant still chooses randomly at stage $2$, and
stage $3$ is random, and the contestant always switches at $4$, then
the winning probability is still ${2 \over 3}$.

Check this by pruning the tree to the first major branch, i.e., by
associating probability $1$ with ``door 1'' at stage $1$.

Of course, this works also for door $2$ and door $3$.
So we have shown that for any choice of prize location, the
contestant's switching strategy still has a ${2 \over 3}$ chance of
winning.

In fact, if the host uses any {\em probability distribution\/} at the
first stage, the contestant still has ${2 \over 3}$ winning chance.
This says the host can't skew the game by any stage 1 strategy.

{\bf Question 3:}  
Stage 1: Random, ${1 \over 3}$ probability of each. \\
Stage 2: Always pick door 1. \\
Stage 3: Random, equal probabilities. \\
Stage 4: Always switch. \\

A: Still ${2 \over 3}$.
Use a different pruning of the tree.

{\bf Question 4:}
Stage 1: A fixed (non-probabilistic) choice. \\
Stage 2: A fixed choice. \\
Stage 3: Random.\\
Stage 4: Switch 

A: There are two situations to consider, which are distinguished by
the kind of {\em knowledge\/} we assume the contestant has.

First, we could assume that the contestant knows the fixed stage 1
choice (if it's always the same, sooner or later, everyone will figure
out what it is).
In this case, the contestant can always win, by first choosing
a wrong door and later switching.
Illustrate this with a pruned tree, pruned at first two stages...

Second, we can assume that the contestant doesn't know the 
fixed stage 1 choice.
This case is modeled in terms of three different trees, one starting
with each of the possible stage $1$ choices.
The contestant's winning guarantee is normally taken to be the {\em
minimum\/} of what he is guaranteed in all three of the trees (that
is, we're taking the worst case probability over three different
possible ``adversaries'' the contestant might be facing).

To be definite, suppose that the contestant's choice at stage 2 is
always door 1.
Then the ``adversary'' that chooses $1$ for the prize location will
always defeat the contestant (with the switching strategy).
So the minimum guarantee will be $0$.

{\bf Question 5:}
Stage $1$: Fixed choice, but unknown to the contestant. \\
Stage $2$: Random, ${1 \over 3}$ probability for each choice. \\
Stage $3$: Host can do whatever he wants. \\
Stage $4$: Switch. 

A: This does not help the host -- it's still ${2 \over 3}$ for the
contestant.
Basically the only choice the host has is at stage $3$, but it doesn't
matter what he does because the switching strategy is guaranteed to
lose anyway.
Trace through the analysis using the tree.

For example, if the host chooses the smallest-numbered possible door
to open at stage 3, then the switching strategy still wins with
probability ${2 \over 3}$.

Many other variations can be considered. 
It's also possible to consider {\em optimality\/} questions, for
instance: 

{\bf Question 6:}
Suppose the host can choose arbitrarily at stage 1,
but chooses randomly at stage 3.
What is the contestant's best strategy?

A: We already know how to achieve ${2 \over 3}$.
We can also prove that this is optimal, as follows:

The contestant strategy can be described by a collection of
eight different parameters:
Use $p_1,p_2,p_3 = 1-p_1-p_2$ for the stage $2$ choices of doors $1$,
$2$, and $3$, respectively.
Then, depending on what happens,
use probability $q_1,q_2,\ldots,q_6$ of switching at stage $4$.
It sounds like there should be 12 situations at stage $4$, but note
that the contestant doesn't know the first stage choice.
This means that the 12 situations can be divided into pairs
that are indistinguishable to the contestant:
for instance, prize behind 1, contestant chooses 1, host opens 2
is indistinguishable from prize behind 3, contestant chooses 1, host opens 2.
So this means that the contestant must use the same probability of
switching, say $q_1$, in both of these cases.

To be definite, suppose the following (choice, opens) pairs yield the
following probabilities of switching:

\noindent
(1,2) yields $q_1$ \\
(1,3) yields $q_2$ \\
(2,3) yields $q_3$ \\
(3,2) yields $q_4$ \\
(2,1) yields $q_5$ \\
(3,1) yields $q_6$ \\

Then brute force analysis says that the probability of winning if
the prize is behind door $1$ is:

$p_1 {1 \over 2} (1-q_1) + p_1 {1 \over 2} (1-q_2) + p_2 q_3 + p_3 q_4$.

The probability if the prize is behind door $2$ is:

$p_1 q_2 + p_2 {1 \over 2} (1-q_5) + p_2 {1 \over 2} (1-q_3) + p_3 q_6$.

And the probability for door $3$ is:

$p_1 q_1 + p_2 q_5 + p_3 {1 \over 2} (1-q_6) + p_3 {1 \over 2} (1-q_4)$.

% Thanks to John Dunagan for the following analysis:

The probability of winning is the minimum of these three, which is at
most equal to the average of these three, which is equal to:

${1 \over 3}
[p_1({1 \over 2}(1-q_1) + {1 \over 2}(1-q_2) + q_2 + q_1)
 + p_2(q_3 + {1 \over 2}(1-q_5) + {1 \over 2}(1-q_3) + q_5)
 + p_3(q_4 + q_6 + {1 \over 2}(1-q_6) + {1 \over 2}(1-q_4))]$,

which is equal to

${1 \over 3}
[p_1(1 + {1 \over 2}(q_1 + q_2))
 + p_2(1 + {1 \over 2}(q_3 + q_5)) 
 + p_3(1 + {1 \over 2}(q_4 + q_6))]$

$\leq {1 \over 3}
[p_1(1 + {1 \over 2}(1 + 1))
 + p_2(1 + {1 \over 2}(1 + 1)) 
 + p_3(1 + {1 \over 2}(1 + 1))]$

$= {2 \over 3}(p_1 + p_2 + p_3) = {2 \over 3}$.

\section{Team Problem}

\begin{problems}

\problem Verify that the set-theoretic probability equations above
``relativize'' to conditional probabilities, e.g., equation~(\ref{a-b})
relativizes to
\[\pr{A-B\mid C}  =  \pr{A \mid C} - \pr{A \intersect B \mid C}.\]


\problem
In the O.J. Simpson trial it was claimed by the defense that the fact that
Simpson battered his wife should be excluded from the evidence as the fraction
of wife batterers that murder their wives is $1/2000$ (Dershowitz Lemma).
In this problem we will condition this probability on the event that
Nicole Simpson is dead and see that we obtain a somewhat different result.

We have the following statistical information (all for 1994):
\begin{itemize}

\item number of women = 125M
\item number of women murdered in 1994 = 6000
\item number of women murdered by spouse in 1994 = 1500
\item number of women with spouse in 1994 = 50M
\item number of women battered by spouse in 1994 = 3M
\end{itemize}

Being murdered by spouse counts as being battered by spouse.  Also, we
assume that the event of being murdered by somebody else is not influenced
by whether or not the woman was being battered by her spouse.

Define the events

$S$ = Woman is murdered by spouse  \\
$M$ = Woman is murdered \\
$B$ = Woman is battered by spouse \\

Use the given information to evaluate the following

\bparts

\ppart  The Dershowitz Lemma can be reformulated as
\[\pr{S \mid B} = 1/2000.\]
Prove it.

\ppart Calculate $\pr{S \mid M}$, i.e., the probability that a murdered
woman was killed by her spouse.

\ppart Calculate $\pr{S \mid B \intersect M}$, i.e., the probability that
a battered, murdered woman is killed by her spouse.

\ppart What can you conclude about Dershowitz' argument that the battery
of a spouse should not be admitted as evidence because of his lemma ?

\eparts

\textbf{Solution:}
\[\pr{S \mid B}= \frac{\pr{S \intersect
B}}{\pr{B}}=\frac{\pr{S}}{\pr{B}}=\frac{1500/125M}{3M/125M}=\frac{1500}{3M}=1/2000.\]

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

Next, note that
\[\pr{S \mid B \intersect M} = \frac{\pr{S \intersect B \intersect M}}{\pr{B \intersect M}} = \frac{\pr{S}}{\pr{M \mid B} \pr{B}}.\]

Then to evaluate $\pr{M \mid B}$, we use the following fact from the
previous problem:
\begin{eqnarray*}
\pr{M \mid B} & = & \pr{M \intersect S \mid B} + \pr{M \intersect \bar{S} \mid B}\\
& = & \pr{S \mid B} + \pr{M \intersect \bar{S}}.
\end{eqnarray*}
In simplifying the second term we used the fact that being murdered by
someone other than the spouse is not influenced by whether the victim was
battered by her spouse or not, namely,
\[\pr{M \intersect \bar{S} \mid B} = \pr{M \intersect \bar{S}}.\]

Now we have
\begin{eqnarray*}
\pr{S \mid B \intersect M} & = & \frac{1500 / 125M}{(\frac{1}{2000} +
\frac{6000- 1500}{125M})\frac{3M}{125M}}\\
& = & \frac{1500 \times 125}{67000 \times 3}\\
 & = & \frac{125}{134}\  =\  0.93.
\end{eqnarray*}

\end{problems}

\end{document}

