\documentclass[twoside,12pt]{article}
\usepackage{light}

%\hidesolutions
\showsolutions

\newcommand{\prcond}[2]{\pr{#1\mid#2}}

\begin{document}

\problemset{9}{April 14, 2005}{Monday, April 25 at 9 PM in Room 32-044}

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

\begin{problem}
There are three coins: a penny, a nickel, and a quarter.  When these
coins are flipped:
%
\begin{itemize}
\item The penny comes up heads with probability $1/3$ and tails with
probability $2/3$.
\item The nickel comes up heads with probability $3/4$ and tails with
probability $1/4$.
\item The quarter comes up heads with probability $3/5$ and tails with
probability $2/5$.
\end{itemize}
%
Assume that the way one coin lands is unaffected by the way the other
coins land.  The goal of this problem is to determine the probability
that an \textit{odd} number of coins come up heads.  For this first
problem, we'll closely follow the four-step procedure for solving
probability problems described in lecture.  \textit{Your solution
should include a tree diagram.}

\bparts

\ppart What is the sample space for this experiment?

\solution{We can regard each outcome as a triple indicating the
orientation of the penny, nickel, and quarter.  For example, the
triple $(H, T, H)$ is the outcome in which the penny is heads, the
nickel is tails, and the quarter is heads.  The sample space is the
set of all such triples:  $\set{H, T}^3$.}

\ppart What subset of the sample space is the event that an odd number
of coins come up heads?

\solution{The event that an odd number of coins come up heads is the
subset:
%
\[
\set{(H, H, H), (H, T, T), (T, H, T), (T, T, H)}
\]
}

\ppart What is the probability of each outcome in the sample space?

\solution{Edges in the tree diagram are labeled with the probabilities
given in the problem statement.  The probability of each outcome is
the product of the probabilities along the corresponding root-to-leaf
path.  The resulting outcome probabilities are noted in the tree
diagram.

\begin{center}
\begin{picture}(300,260)(0,-30)
%\put(0,-30){\dashbox(300,260){}}
\put(180,0){\line(-4,1){60}}
\put(180,30){\line(-4,-1){60}}
\put(180,60){\line(-4,1){60}}
\put(180,90){\line(-4,-1){60}}
\put(180,120){\line(-4,1){60}}
\put(180,150){\line(-4,-1){60}}
\put(180,180){\line(-4,1){60}}
\put(180,210){\line(-4,-1){60}}
\put(120,15){\line(-2,1){60}}
\put(120,75){\line(-2,-1){60}}
\put(120,135){\line(-2,1){60}}
\put(120,195){\line(-2,-1){60}}
\put(60,45){\line(-1,1){60}}
\put(60,165){\line(-1,-1){60}}
\put(30,150){\makebox(0,0){$H$}}
\put(30,60){\makebox(0,0){$T$}}
\put(100,195){\makebox(0,0){$H$}}
\put(100,135){\makebox(0,0){$T$}}
\put(100,75){\makebox(0,0){$H$}}
\put(100,15){\makebox(0,0){$T$}}
\put(170,-5){\makebox(0,0){$T$}}
\put(170,35){\makebox(0,0){$H$}}
\put(170,55){\makebox(0,0){$T$}}
\put(170,95){\makebox(0,0){$H$}}
\put(170,115){\makebox(0,0){$T$}}
\put(170,155){\makebox(0,0){$H$}}
\put(170,175){\makebox(0,0){$T$}}
\put(170,215){\makebox(0,0){$H$}}
\put(30,40){\makebox(0,0){\text{penny}}}
\put(90,0){\makebox(0,0){\text{nickel}}}
\put(150,-20){\makebox(0,0){\text{quarter}}}
\put(210,-20){\makebox(0,0){\text{odd?}}}
\put(210,210){\makebox(0,0){$\times$}}
\put(210,120){\makebox(0,0){$\times$}}
\put(210,60){\makebox(0,0){$\times$}}
\put(210,30){\makebox(0,0){$\times$}}
\put(270,-20){\makebox(0,0){\text{prob.}}}
\put(35,120){\makebox(0,0){$1/3$}}
\put(35,90){\makebox(0,0){$2/3$}}
\put(75,185){\makebox(0,0){$3/4$}}
\put(75,145){\makebox(0,0){$1/4$}}
\put(75,65){\makebox(0,0){$3/4$}}
\put(75,25){\makebox(0,0){$1/4$}}
\put(140,210){\makebox(0,0){$3/5$}}
\put(140,180){\makebox(0,0){$2/5$}}
\put(140,150){\makebox(0,0){$3/5$}}
\put(140,120){\makebox(0,0){$2/5$}}
\put(140,90){\makebox(0,0){$3/5$}}
\put(140,60){\makebox(0,0){$2/5$}}
\put(140,30){\makebox(0,0){$3/5$}}
\put(140,0){\makebox(0,0){$2/5$}}
\put(270,210){\makebox(0,0){$9/60$}}
\put(270,180){\makebox(0,0){$6/60$}}
\put(270,150){\makebox(0,0){$3/60$}}
\put(270,120){\makebox(0,0){$2/60$}}
\put(270,90){\makebox(0,0){$18/60$}}
\put(270,60){\makebox(0,0){$12/60$}}
\put(270,30){\makebox(0,0){$6/60$}}
\put(270,0){\makebox(0,0){$4/60$}}
\end{picture}
\end{center}
}

\ppart What is the probability that an odd number of coins come up
heads?

\solution{The probability of an event is the sum of the probabilities
of the outcomes in that event.  In this case:
%
\begin{align*}
\lefteqn{\pr{\text{odd number of heads}}}
\hspace{1in} \\
    & = \pr{\set{(H, H, H), (H, T, T), (T, H, T), (T, T, H)}} \\
    & = \pr{(H, H, H)} +
        \pr{(H, T, T)} +
        \pr{(T, H, T)} +
        \pr{(T, T, H)} \\
    & = \frac{9}{60} + \frac{2}{60} + \frac{12}{60} + \frac{6}{60} \\
    & = \frac{29}{60}
\end{align*}
}

\eparts

\end{problem}

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

\begin{problem}
Professor Plum, Mr. Green, and Miss Scarlet are all plotting to shoot
Colonel Mustard.  If one of these three has both an
\textit{opportunity} and the \textit{revolver}, then that person shoots
Colonel Mustard.  Otherwise, Colonel Mustard escapes.  Exactly one of
the three has an \textit{opportunity} with the following
probabilities:
%
\begin{align*}
\pr{\text{Plum has opportunity}} & = 1 / 6 \\
\pr{\text{Green has opportunity}} & = 2 / 6 \\
\pr{\text{Scarlet has opportunity}} & = 3 / 6
\end{align*}
%
Exactly one has the \textit{revolver} with the following
probabilities, regardless of who has an opportuntity:
%
\begin{align*}
\pr{\text{Plum has revolver}} & = 4 / 8 \\
\pr{\text{Green has revolver}} & = 3 / 8 \\
\pr{\text{Scarlet has revolver}} & = 1 / 8
\end{align*}

\bparts

\ppart Draw a tree diagram for this problem.  Indicate edge and
outcome probabilities.

\solution{
\begin{center}
\begin{picture}(120,270)(0,-30)
\put(0,120){\line(2,3){60}}
\put(0,120){\line(1,0){60}}
\put(0,120){\line(2,-3){60}}
\put(60,30){\line(2,-1){60}}
\put(60,30){\line(1,0){60}}
\put(60,30){\line(2,1){60}}
\put(60,120){\line(2,-1){60}}
\put(60,120){\line(1,0){60}}
\put(60,120){\line(2,1){60}}
\put(60,210){\line(2,-1){60}}
\put(60,210){\line(1,0){60}}
\put(60,210){\line(2,1){60}}
\put(30,180){\makebox(0,0){P}}
\put(30,130){\makebox(0,0){G}}
\put(30,60){\makebox(0,0){S}}
\put(110,245){\makebox(0,0){P}}
\put(110,220){\makebox(0,0){G}}
\put(110,175){\makebox(0,0){S}}
\put(110,155){\makebox(0,0){P}}
\put(110,130){\makebox(0,0){G}}
\put(110,85){\makebox(0,0){S}}
\put(110,65){\makebox(0,0){P}}
\put(110,40){\makebox(0,0){G}}
\put(110,-5){\makebox(0,0){S}}
\put(45,160){\makebox(0,0){$1/6$}}
\put(30,110){\makebox(0,0){$2/6$}}
\put(45,80){\makebox(0,0){$3/6$}}
\put(85,235){\makebox(0,0){$4/8$}}
\put(110,200){\makebox(0,0){$3/8$}}
\put(85,185){\makebox(0,0){$1/8$}}
\put(85,145){\makebox(0,0){$4/8$}}
\put(110,110){\makebox(0,0){$3/8$}}
\put(85,95){\makebox(0,0){$1/8$}}
\put(85,55){\makebox(0,0){$4/8$}}
\put(110,20){\makebox(0,0){$3/8$}}
\put(85,5){\makebox(0,0){$1/8$}}
\put(30,15){\makebox(0,0){\text{opportunity}}}
\put(90,-20){\makebox(0,0){\text{revolver}}}
\put(150,-20){\makebox(0,0){\text{prob.}}}
\put(150,240){\makebox(0,0){$4/48$}}
\put(150,210){\makebox(0,0){$3/48$}}
\put(150,180){\makebox(0,0){$1/48$}}
\put(150,150){\makebox(0,0){$8/48$}}
\put(150,120){\makebox(0,0){$6/48$}}
\put(150,90){\makebox(0,0){$2/48$}}
\put(150,60){\makebox(0,0){$12/48$}}
\put(150,30){\makebox(0,0){$9/48$}}
\put(150,0){\makebox(0,0){$3/48$}}
\end{picture}
\end{center}
}

\ppart What is the probability that Colonel Mustard is shot?

\solution{Denote each outcome with a pair indicating who has the
opportunity and who has the revolver.  In this notation, the event
that Colonel Mustard is shot consists of all outcomes where a single
person has both:
%
\[
\set{(P,P),(G,G),(S,S)}
\]n
%
The probability of this event is the sum of the outcome probabilities:
%
\begin{align*}
\pr{\set{(P,P),(G,G),(S,S)}}
    & = \pr{(P,P)} + \pr{(G,G)} + \pr{(S,S)} \\
    & = 4/48 + 6/48 + 3/48 \\
    & = 13/48
\end{align*}
}

\ppart What is the probability that Colonel Mustard is shot, given
that Miss Scarlet does not have the revolver?

\solution{Let $S$ be the event that Colonel Mustard is shot, and let
$N$ be the event that Miss Scarlet does \textit{not} have the
revolver.  The solution is:
%
\begin{align*}
\pr{S \mid N}
    & = \frac{\pr{S \cap N}}{\pr{N}} \\
    & = \frac{\pr{(P,P),(G,G)}}{\pr{(P,P),(P,G),(G,P),(G,G),(S,P),(S,G)}} \\
    & = \frac{\frac{4}{48}+\frac{6}{48}}
             {\frac{4}{48}+\frac{3}{48}+\frac{8}{48}+
              \frac{6}{48}+\frac{12}{48}+\frac{9}{48}} \\
    & = \frac{5}{21}
\end{align*}
}

\ppart What is the probability that Mr. Green had an opportunity,
given that Colonel Mustard was shot?

\solution{Let $G$ be the event that Mr. Green has an opportunity, and
again let $S$ be the event that Colonel Mustard is shot.  Then the
solution is:
%
\begin{align*}
\pr{G \mid S}
    & = \frac{\pr{G \cap S}}{\pr{S}} \\
    & = \frac{\pr{(G,G)}}{\pr{(P,P),(G,G),(S,S)}} \\
    & = \frac{\frac{6}{48}}
             {\frac{4}{48}+\frac{6}{48}+\frac{3}{48}} \\
    & = \frac{6}{13}
\end{align*}
}

\eparts

\end{problem}

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

\begin{problem}
There are three prisoners in a maximum-security prison for fictional
villains: the Evil Wizard Voldemort, the Dark Lord Sauron, and Little
Bunny Foo-Foo.  The parole board has declared that it will release two
of the three, chosen uniformly at random, but has not yet released
their names.  Naturally, Sauron figures that he will be released to
his home in Mordor, where the shadows lie, with probability
$\frac{2}{3}$.

A guard offers to tell Sauron the name of one of the other prisoners
who will be released (either Voldemort or Foo-Foo).  However, Sauron
declines this offer.  He reasons that if the guard says, for example,
``Little Bunny Foo-Foo will be released'', then his own probability of
release will drop to $\frac{1}{2}$.  This is because he will then know
that either he or Voldemort will also be released, and these two
events are equally likely.

Using a tree diagram and the four-step method, either prove that the
Dark Lord Sauron has reasoned correctly or prove that he is wrong.
Assume that if the guard has a choice of naming either Voldemort or
Foo-Foo (because both are to be released), then he names one of the
two uniformly at random.
\end{problem}

\solution{Sauron has reasoned incorrectly.  In order to understand his
error, let's begin by working out the sample space, noting events of
interest, and computing outcome probabilities:
%
\begin{center}
\begin{picture}(360,175)(0,-40)
%\put(0,-40){\dashbox(360,175){}} % bounding box
\put(0,60){\line(1,1){60}}
\put(0,60){\line(1,0){60}}
\put(0,60){\line(1,-1){60}}
\put(30,-10){\makebox(0,0){released}}
\put(60,120){\line(1,0){60}}
\put(60,60){\line(2,1){60}}
\put(60,60){\line(2,-1){60}}
\put(60,0){\line(1,0){60}}
\put(90,-25){\makebox(0,0){guard says}}
\put(11,90){\makebox(0,0){$F,S$}}
\put(40,68){\makebox(0,0){$F,V$}}
\put(11,30){\makebox(0,0){$V,S$}}
\put(52,96){\makebox(0,0){$1/3$}}
\put(40,50){\makebox(0,0){$1/3$}}
\put(52,24){\makebox(0,0){$1/3$}}
\put(90,128){\makebox(0,0){$F$}}
\put(90,90){\makebox(0,0){$F$}}
\put(90,30){\makebox(0,0){$V$}}
\put(90,-10){\makebox(0,0){$V$}}
\put(90,110){\makebox(0,0){$1$}}
\put(102,70){\makebox(0,0){$1/2$}}
\put(102,50){\makebox(0,0){$1/2$}}
\put(90,8){\makebox(0,0){$1$}}
\put(150,120){\makebox(0,0){$1/3$}}
\put(150,90){\makebox(0,0){$1/6$}}
\put(150,30){\makebox(0,0){$1/6$}}
\put(150,0){\makebox(0,0){$1/3$}}
\put(150,-20){\makebox(0,0){prob.}}
\put(210,120){\makebox(0,0){$\times$}}
\put(210,90){\makebox(0,0){$\times$}}
\put(210,30){\makebox(0,0){}}
\put(210,0){\makebox(0,0){}}
\put(210,-25){\makebox(0,0){\shortstack{guard says\\"Foo-foo''}}}
\put(270,120){\makebox(0,0){$\times$}}
\put(270,90){\makebox(0,0){$\times$}}
\put(270,30){\makebox(0,0){$\times$}}
\put(270,0){\makebox(0,0){}}
\put(270,-25){\makebox(0,0){\shortstack{Foo-foo\\released}}}
\put(330,120){\makebox(0,0){$\times$}}
\put(330,90){\makebox(0,0){}}
\put(330,30){\makebox(0,0){}}
\put(330,0){\makebox(0,0){$\times$}}
\put(330,-25){\makebox(0,0){\shortstack{Sauron\\released}}}
\end{picture}
\end{center}

Define the events $S$, $F$, and ``$F$'' as follows:
%
\begin{align*}
\text{``$F$''} & = \text{Guard says Foo-Foo is released} \\
F & = \text{Foo-Foo is released} \\
S & = \text{Sauron is released}
\end{align*}
%
The outcomes in each of these events are noted in the tree diagram.

Sauron's error is in failing to realize that the event $F$ (Foo-foo
will be released) is different from the event ``$F$'' (the guard
\textit{says} Foo-foo will be released).  In particular, the
probability that Sauron is released, given that Foo-foo is released,
is indeed $1/2$:
%
\begin{align*}
\pr{S \mid F}
    & = \frac{\pr{S \cap F}}{\pr{F}} \\
    & = \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}+\frac{1}{6}} \\
    & = \frac{1}{2}
\end{align*}
%
But the probability that Sauron is released given that the guard
merely \textit{says so} is still $2/3$:
%
\begin{align*}
\pr{S \mid \text{``$F$''}}
    & = \frac{\pr{S \cap \text{``$F$''}}}{\pr{\text{``$F$''}}} \\
    & = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{6}} \\
    & = \frac{2}{3}
\end{align*}
%
So Sauron's probability of release is actually unchanged by the
guard's statement.}

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

\begin{problem}
You shuffle a deck of cards and deal your friend a 5-card hand.

\bparts

\ppart Suppose your friend says, ``I have the ace of spades.''  What
is the probablity that she has another ace?

\solution{The sample space for this experient is the set of all 5-card
hands.  All outcomes are equally likely, so the probability of each
outcome is $1 / \binom{52}{5}$.  Let $S$ be the event that your friend
has the ace of spades, and let $A$ be the event that your friend has
another ace.  Our objective is to compute:
%
\[
\pr{A \mid S} = \frac{\pr{A \cap S}}{\pr{S}}
\]
%
The number of hands containing the ace of spades is equal to the
number of ways to select 4 of the remaining 51 cards.  Therefore:
%
\[
\pr{S} = \frac{\binom{51}{4}}{\binom{52}{5}}
\]
%
The number of hands containing the ace of spades and at least one more
ace is:
%
\[
\size{A \cap S} = \binom{3}{1} \binom{48}{3} + \binom{3}{2}
\binom{48}{2} + \binom{3}{3} \binom{48}{1}
\]
%
Here the first term counts the number of hands with one additional
ace, since there are $\binom{3}{1}$ ways to choose the extra ace and
$\binom{48}{3}$ ways to choose the other cards.  Similarly, the second
term counts hands with two additional aces, and the last term counts
hands with all three remaining aces.  In probability terms, we have:
%
\[
\pr{A \cap S} = \frac{\binom{3}{1} \binom{48}{3} + \binom{3}{2} \binom{48}{2} + \binom{3}{3} \binom{48}{1}}{\binom{52}{5}}
\]
%
Substituting these results into our original equation gives the solution:
%
\[
\pr{A \mid S} = \frac{\binom{3}{1} \binom{48}{3} + \binom{3}{2} \binom{48}{2} + \binom{3}{3} \binom{48}{1}}{\binom{51}{4}} = 0.2214\ldots
\]
}

\ppart Suppose your friend says, ``I have an ace.''  What is the
probability that she has another ace?

\solution{The sample space and outcome probabilities are the same as
before.  Let $L$ be the event that your friend has at least one ace,
and $M$ be the event that your friend has more than one ace.  Our goal
it to compute:
%
\[
\pr{M \mid L} = \frac{\pr{M \cap L}}{\pr{L}} = \frac{\pr{M}}{\pr{L}}
\]
%
The second equality holds because your friend surely at least one ace
if she has more than one; that is, $M \subseteq L$.  The probability
that your friend has at least one ace is:
%
\[
\pr{L} = \frac{
  \binom{4}{1}\binom{48}{4} +
  \binom{4}{2}\binom{48}{3} +
  \binom{4}{3}\binom{48}{2} +
  \binom{4}{4}\binom{48}{1}}{\binom{52}{5}}
\]
%
The first term counts hands with a single ace, since there are
$\binom{4}{1}$ ways to choose the ace and $\binom{48}{4}$ ways to
choose the remaining four cards.  The remaining terms are similar.
The probability that your friend has more than one ace is:
%
\[
\pr{M} = \frac{
  \binom{4}{2}\binom{48}{3} +
  \binom{4}{3}\binom{48}{2} +
  \binom{4}{4}\binom{48}{1}}{\binom{52}{5}}
\]
%
Plugging these results into the original equation gives:
%
\[
\pr{M \mid L} =
  \frac{\binom{4}{2}\binom{48}{3} +
        \binom{4}{3}\binom{48}{2} +
        \binom{4}{4}\binom{48}{1}}
       {\binom{4}{1}\binom{48}{4} +
        \binom{4}{2}\binom{48}{3} +
        \binom{4}{3}\binom{48}{2} +
        \binom{4}{4}\binom{48}{1}} = 0.12285\ldots
\]
}

\ppart Are your answers to (a) and (b) the same?  Explain why.

\solution{The answers are different.  There are four aces, so there
are sixteen different subsets of aces that your friend could have.
%
\begin{itemize}
\item If your friend says, ``I have the ace of spades'', then
\textit{eight} of these subsets are ruled out: those not containing
the ace of spades.
\item However, if your friend says, ``I have an ace'', then only
\textit{one} subset is ruled out: the subset containing no aces.
\end{itemize}
%
Thus, the probability that your friend has a second ace is different
in these two cases, because we are conditioning on two very different
events!}

\eparts

\end{problem}

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

\iffalse
\begin{problem}
Using only a slightly bent coin, three roommates must decide who has
to scrub the bathtub.  They hit upon the following procedure.
%
\begin{enumerate}

\item Flip the coin three times.

\item If the coin lands the same way every time, then go to step 1.

\item Otherwise, one of the flips lands differently from the other
two.  If the $n$-th toss is the different one, then the $n$-th
roommate must clean the tub.  For example, if the last three flips are
$THT$, then the second roommate gets friendly with soapscum.

\end{enumerate}
%
Suppose that the slightly-bent coin lands heads-up with probability
$p$, where $0 < p < 1$.

(For this problem, the complete tree diagram is infinite.  However, a
tree

\bparts

\ppart What is the probability that the branch back to step 1 is
taken?

\solution{$p^3 + (1-p)^3$}

\ppart If step 3 is reached, what is the probability that two of the
last three flips were heads?

\solution{$p$}

\ppart Compute the probability that the second roommate must clean the
tub.

\eparts

(For this problem, the tree diagram is actually \textit{infinite}.
However, drawing enough of the tree diagram to see a pattern may be
helpful.  To compute the probability of an infinite envent, you must
sum an infinite number of outcome probabilities.)

\end{problem}
\fi

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

\begin{problem} (From an old final exam.)  \textit{Finalphobia} is a rare
disease in which the victim has the delusion that he or she is being
subjected to an intense mathematical examination.
%
\begin{itemize}
\item A person selected uniformly at random has finalphobia with probability
$1/100$.
\item A person with finalphobia has shaky hands with probability $9/10$.
\item A person without finalphobia has shaky hands with probability $1/20$.
\end{itemize}
%
What is the probablility that a person selected uniformly at random
has finalphobia, given that he or she has shaky hands?

\solution{
Let $F$ be the event that the randomly-selected person has
finalphobia, and let $S$ be the event that he or she has shaky hands.
A tree diagram is worked out below:
%
\begin{center}
\includegraphics{finalphobia}
\end{center}
%
The probability that a person has finalphobia, given that he or she
has shaky hands is:
%
\begin{align*}
\pr{F \mid S}
    & = \frac{\pr{F \cap S}}{\pr{S}} \\
    & = \frac{9/1000}{9/1000 + 99/2000} \\
    & = \frac{18}{18+99} \\
    & = \frac{18}{117}
\end{align*}
}

\end{problem}

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

\begin{problem}
Outside of their hum-drum duties as 6.042 TAs, Ishan is trying to
learn to levitate using only intense concentration and Grant is
launching a ``Wang 2008'' presidential campaign.  Suppose that Ishan's
probability of levitating is $1/6$, Grant's chance of becoming
president is $1/4$, and the success of one does not alter the other's
chances.

\bparts

\ppart If at least one of them succeeds, what is the probability that
Ishan learns to levitate?

\solution{Let $L$ be the event that Ishan learns to levitate, and let
$P$ be the event that Grant becomes president.  We can work out the
desired probability as follows:
%
\begin{align*}
\pr{L \mid (L \cup P)}
	& = \frac{\pr{L \cap (L \cup P)}}{\pr{L \cup P}} \\
	& = \frac{\pr{L}}{1 - \pr{\overline{L} \cap \overline{P}}} \\
	& = \frac{1/6}{1 - (1 - 1/6)(1 - 1/4)} \\
	& = \frac{4}{9}
\end{align*}
%
The first step uses the definition of conditional probability.  In the
second step, we rewrite both the top and bottom of the fraction using
set identities.  Then we substitute in the given probability and
simplify.}

\ppart If at most one of them succeeds, what is the probability that
Grant becomes the president of the United States?

\solution{
Define events $L$ and $P$ as before.
%
\begin{align*}
\pr{P \mid (\overline{L} \cup \overline{P})}
	& = \frac{\pr{P \cap (\overline{L} \cup \overline{P})}}
		{\pr{\overline{L} \cup \overline{P}}} \\
	& = \frac{\pr{P \cap \overline{L}}}
		{1 - \pr{L \cap P}} \\
	& = \frac{(1/4) \cdot (5/6)}{1 - (1/6) \cdot (1/4)} \\
	& = \frac{5}{23}
\end{align*}
}

\ppart If exactly one of them succeeds, what is the probability that
it is Ishan?

\solution{
\begin{align*}
\pr{L \mid ((L \cap \overline{P}) \cup (\overline{L} \cap P))}
	& = \frac{\pr{L \cap \overline{P}}}
		{\pr{((L \cap \overline{P}) \cup
		(\overline{L} \cap P))}} \\
	& = \frac{(1/6) \cdot (3/4)}
		{(1/6) \cdot (3/4) + (5/6) \cdot (1/4)} \\
	& = \frac{3}{8}
\end{align*}
}

\eparts
\end{problem}

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

\begin{problem}
My computer came with the game \textit{Minesweeper}.  In this game,
there is an $8 \times 8$ grid of squares.  Ten randomly-selected
squares contain mines, and all configurations of mines are equally
likely.

\bparts

\ppart What is the sample space for this game and what is the
probability of each outcome?

\solution{The sample space consists of all arrangements of 10 mines on
the $8 \times 8$ board.  All such configurations have the same
probability, which therefore must be $1 / \binom{64}{10}$.}

\ppart After I made a few moves, the game board looked like this:
%
\begin{center}
\begin{picture}(200,160)(-20,0)
% \put(-20,0){\dashbox(200,160){}} % bounding box
\multiput(0,0)(20,0){9}{\line(0,1){160}}
\multiput(0,0)(0,20){9}{\line(1,0){160}}
\put(70,110){\makebox(0,0){1}}
\put(90,110){\makebox(0,0){1}}
\put(110,110){\makebox(0,0){3}}
\end{picture}
\end{center}
%
Now I know that the three numbered squares do \textit{not} contain
mines.  Furthermore, each number indicates how many squares adjacent
to that number \textit{do} contain mines.  (Two squares are adjacent
if they share an edge or a corner.)  Let $E \subseteq S$ be the event
that the configuration of mines is consistent with this numbering.
Describe all outcomes in the event $E$.

\solution{Here is the board with some squares labeled for discussion:
%
\begin{center}
\begin{picture}(160,160)
\multiput(0,0)(20,0){9}{\line(0,1){160}}
\multiput(0,0)(0,20){9}{\line(1,0){160}}
\put(70,110){\makebox(0,0){1}}
\put(90,110){\makebox(0,0){1}}
\put(110,110){\makebox(0,0){3}}
\put(50,90){\makebox(0,0){$a$}}
\put(50,110){\makebox(0,0){$a$}}
\put(50,130){\makebox(0,0){$a$}}
\put(70,90){\makebox(0,0){$b$}}
\put(70,130){\makebox(0,0){$b$}}
\put(90,90){\makebox(0,0){$c$}}
\put(90,130){\makebox(0,0){$c$}}
\put(110,90){\makebox(0,0){$d$}}
\put(110,130){\makebox(0,0){$d$}}
\put(130,90){\makebox(0,0){$e$}}
\put(130,110){\makebox(0,0){$e$}}
\put(130,130){\makebox(0,0){$e$}}
\end{picture}
\end{center}

We can divide all the outcomes in $E$ into three groups based on the
distribution of mines among the squares labeled $a$, $b$, $c$, $d$,
and $e$ and the 49 blank squares:
%
\begin{center}
\begin{tabular}{ccccccl}
\multicolumn{6}{c}{\# of mines} \\\
a & b & c & d & e & other & \# of configurations\\[1ex]
1 & 0 & 0 & 1 & 2 & 6 & $\binom{3}{1} \binom{2}{0} \binom{2}{0} \binom{2}{1} \binom{3}{2} \binom{49}{6}$ \\[1ex]
0 & 1 & 0 & 0 & 3 & 6 & $\binom{3}{0} \binom{2}{1} \binom{2}{0} \binom{2}{0} \binom{3}{3} \binom{49}{6}$ \\[1ex]
0 & 0 & 1 & 0 & 2 & 7 & $\binom{3}{0} \binom{2}{0} \binom{2}{1} \binom{2}{0} \binom{3}{2} \binom{49}{7}$
\end{tabular}
\end{center}
}

\ppart On the next move, I must click an unnumbered square.  If that
square contains a mine, I lose the game!  The squares marked $x$ and
$y$ in the diagram below look like reasonable choices.
%
\begin{center}
\begin{picture}(200,160)(-20,0)
% \put(-20,0){\dashbox(200,160){}} % bounding box
\multiput(0,0)(20,0){9}{\line(0,1){160}}
\multiput(0,0)(0,20){9}{\line(1,0){160}}
\put(70,110){\makebox(0,0){1}}
\put(90,110){\makebox(0,0){1}}
\put(110,110){\makebox(0,0){3}}
\put(-10,150){\vector(4,-1){80}}
\put(-15,150){\makebox(0,0){x}}
\put(170,150){\vector(-4,-1){80}}
\put(175,150){\makebox(0,0){y}}
\end{picture}
\end{center}
%
What is the probability that there is a mine at the square marked $x$?

\solution{If there is a mine at square $x$, then the state of all the
other squares next to numbers is complete determined and 6 of the
other 49 squares are mined.  Therefore, the probability that $x$ is
mined is:
%
\[
\pr{\text{x mined} \mid E} = \frac{\pr{\text{x mined} \cap E}}{\pr{E}} = \frac{\binom{49}{6}}{\size{E}} = 0.0175\ldots
\]}

\ppart What is the probability that there is a mine at the square
marked $y$?

\solution{If there is a mine at square $y$, then two of the three
squares marked $e$ are mined as well as 7 of the remaining 49 squares.
Thus, the probability that there is a mine at $y$ is:
%
\[
\pr{\text{y mined} \mid E} = \frac{\pr{\text{y mined} \cap E}}{\pr{E}} = \frac{\binom{3}{2}\binom{49}{7}}{\size{E}} = 0.324\ldots
\]
%
Thus, a mine at $y$ is 18+ times more likely than a mine $x$!}

\eparts

\noindent (You may leave your answers in symbolic form, but the
numerical answers are surely more interesting!)

\end{problem}

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

\end{document}
