%  Homework assignment 7 solutions
%  21-110: Problem Solving in Recreational Mathematics, spring 2010
%  Assigned: Monday, April 5, 2010. Due: Wednesday, April 14, 2010.
%  Brian Kell <bkell@cmu.edu>

\pdfpagewidth=8.5truein \pdfpageheight=11truein
\hsize=6truein \vsize=9truein
\pdfhorigin=1.25truein \pdfvorigin=1truein

% MetaPost stuff
\input supp-pdf
\def\fig#1{\convertMPtoPDF{#1}{1}{1}}
\def\midfig#1{{%
	\setbox0=\hbox{\fig{#1}}%
	\dimen0=0.5\ht0%
	\advance\dimen0 by-0.3\ht\strutbox
	\lower\dimen0\box0%
}}

% AMSFonts
\input amssym.tex

\font\titlefont=cmbx12
\font\smfont=cmr8

\headline={\hfil}
\footline={\hfil Page~\folio\hfil}
\topskip=20pt plus2pt

\newcount\pno  \pno=0
\def\problem{\goodbreak\advance\pno by1 \noindent{\bf Problem~\the\pno.} }
\def\problemno#1. {\goodbreak\noindent{\bf Problem~\hbox{#1}.} }
\def\noskiphint#1\par{\nobreak{\hyphenpenalty=10000\noindent{\it Hint:\/}\enspace#1\par}}
\def\hint#1\par{\nobreak\medskip\noskiphint#1\par}
\def\itemhint#1\par{\nobreak\smallskip\item{}{\it Hint:\/}\enspace#1\par}
\def\solution{\nobreak\medskip\noindent{\bf Solution.} }
\def\part#1{\goodbreak\medskip\indent\llap{#1\enspace}\ignorespaces}
\def\firstpart#1{\nobreak\medskip\indent\llap{#1\enspace}\ignorespaces}

% Pilfered from an example on page 106 of the TeXbook
\def\qed{{\unskip\nobreak\hfil\penalty50
	\hskip1em\hbox{}\nobreak\hfil\hbox{\vrule\vbox to6pt{\hrule\hbox to5.2pt{\hss}\vfil\hrule}\vrule}
	\parfillskip=0pt \finalhyphendemerits=0 \par}}

\def\ipmod#1{\;({\rm mod}\,\,#1)}  % looks better inline than \pmod does
\def\clap#1{\hbox to 0pt{\hss$\scriptstyle{#1}$\hss}}
\def\lcm{\mathop{\rm lcm}}  \def\gcd{\mathop{\rm gcd}}
\def\labs{\mathopen|}  \def\rabs{\mathclose|}  \def\abs#1{\labs{#1}\rabs}
\def\N{{\mathord{\Bbb N}}}  \def\Z{{\mathord{\Bbb Z}}}
\def\frac#1#2{{\textstyle{{#1}\over{#2}}}}

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

\centerline{\titlefont 21-110: Problem Solving in Recreational Mathematics}
\smallskip
\centerline{\bf Homework assignment~7 solutions}

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem An urn contains five red balls and three yellow balls. Two balls are
drawn from the urn at random, without replacement.

\nobreak\smallskip

\item{(a)} In this scenario, what is the experiment? What is the sample space?

\nobreak\smallskip

\item{(b)} What is the probability that the first ball drawn is red?

\nobreak\smallskip

\item{(c)} What is the probability that at least one of the two balls drawn is
red?

\nobreak\smallskip

\item{(d)} What is the (conditional) probability that the second ball drawn is
red, given that the first ball drawn is red?

\solution

\firstpart{(a)} The experiment is the drawing of two balls from the urn without
replacement. The sample space is the set of possible outcomes, of which there
are four: drawing two red balls; drawing two yellow balls; drawing a red ball
first, and then a yellow ball; and drawing a yellow ball first, and then a red
ball. One way to denote the sample space is in set notation, abbreviating the
colors red and yellow:
$$\hbox{sample space}=\{\rm RR,YY,RY,YR\}.$$
Note that these four outcomes are {\it not\/} equally likely.

We can also represent the experiment and the possible outcomes in a probability
tree diagram, as shown below. Note in particular the probabilities given for
the second ball. For example, if the first ball is red, then four out of the
remaining seven balls are red, so the probability that the second ball is red
is~$4/7$ (and the probability that it is yellow is~$3/7$). On the other hand,
if the first ball is yellow, then five out of the remaining seven balls are
red, so the probability that the second ball is red is~$5/7$ (and the
probability that it is yellow is~$2/7$). The probabilities of each of the four
outcomes can be computed by multiplying the probabilities along the branches
leading to the outcomes.
$$\fig{homework-7-sol-ball-tree.mps}$$

\part{(b)} Before the first draw, five out of the eight balls in the urn are
red. Each ball is equally likely to be drawn. So the probability that the first
ball drawn is red is~$5/8$.

\part{(c)} The event that at least one of the two balls is red contains three
outcomes: RR,~RY, and~YR. Since we know the probabilities of all of these
outcomes, we can find the probability of this event by adding the probabilities
of the individual outcomes. So the probability that at least one of the two
balls is red is $5/14+15/56+15/56=25/28$.

Alternatively, we can observe that the event that at least one of the two balls
is red is the complement of the event that both balls are yellow. The
probability that both balls are yellow is~$3/28$, so the probability that it is
{\it not\/} the case that both balls are yellow (i.e., the probability that at
least one ball is red) is $1-3/28=25/28$.

\part{(d)} Suppose the first ball drawn is red. Then there are seven remaining
balls, and four of them are red. So the conditional probability that the second
ball is red, given that the first ball is red, is~$4/7$. \qed

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\cents{\hbox{\rm\rlap/c}}

\problem Suppose you and a friend play a game. Two standard, fair, six-sided
dice are thrown, and the numbers appearing on the dice are multiplied together.
If this product is even, your friend gives you a quarter, but if this product
is odd, you must give your friend one dollar.

\nobreak\smallskip

\item{(a)} What is the expected value of this game for you? Round to the
nearest cent.

\nobreak\smallskip

\item{(b)} What is the expected value of this game for your friend? Round to
the nearest cent.

\nobreak\smallskip

\item{(c)} Is this game fair? Why or why not?

\solution We begin by drawing a table of the possible dice throws. The
36~outcomes represented by the squares of the grid below are all equally
likely. The shaded squares represent outcomes in which the product of the
numbers on the dice is even, and the unshaded squares represent outcomes in
which this product is odd.
$$\fig{homework-7-sol-dice-table.mps}$$

\part{(a)} From the table above, we see that 27 of the 36~equally likely
outcomes yield an even product, so the probability that the product is even
is~$27/36$, or~$3/4$. Similarly, the probability that the product is odd
is~$9/36$, or~$1/4$. You will win~25\cents\ if the product of the numbers on
the dice is even, but you will lose~\$1 if the product is odd. So there is a
$3/4$~probability that you will gain~25\cents\ and a $1/4$~probability that you
will lose~100\cents\ (that~is, gain~$-100\cents$). So the expected value of the
game for you is
$$\frac34(25\cents)+\frac14(-100\cents)=-6.25\cents.$$
Rounded to the nearest cent, this is~$-6\cents$. This means that you should
expect to lose about~6\cents, on average, each time you play this game.

\part{(b)} Any money that you win in this game is money that your friend loses,
and any money that your friend wins is money that you lose. (Mathematicians
call this a {\it zero-sum game}.) Therefore, if the expected value of the game
for you is~$-6\cents$, the expected value of the game for your friend
is~$6\cents$. Your friend should expect to win about~6\cents, on average, each
time the game is played.

\part{(c)} This game is not fair, because the expected value of the game is not
exactly zero. \qed

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Suppose you roll a fair six-sided die six times and record the results
of the rolls in order. What is the probability that you roll a~``3'' at least
five times in a row?

\solution There are essentially two ways in which this event can occur: either
the first five rolls are all~``3,'' or the last five rolls are all~``3.'' Let's
give names to these events; let $A$ be the event that the first five rolls are
all~``3,'' and let $B$ be the event that the last five rolls are all~``3.'' The
question asks for the probability that either of these events occur, which is
the probability of the event~$A\cup B$.

How many outcomes are there in all? In other words, what is the cardinality of
the sample space? Each roll of the die has six possible outcomes, so, by the
multiplication principle, in six rolls of the die there are~$6^6$, or~46,656,
possible outcomes (because the order of the rolls is important). These outcomes
are all equally likely.

How many outcomes are in the event~$A$? The outcomes in the event~$A$ are those
that look like ``3,~3,~3,~3,~3,~?,'' where the question mark can be replaced by
any number from 1 through~6. So the cardinality of~$A$ is~6.

Likewise, the outcomes in the event~$B$ are those that look like
``?,~3,~3,~3,~3,~3,'' where the question mark can be replaced with any number
from 1 through~6. So the cardinality of~$B$ is also~6.

How many outcomes are in both $A$ and~$B$? In other words, what is the
cardinality of~$A\cap B$? There is only one such outcome, namely,
``3,~3,~3,~3,~3,~3.'' So $\abs{A\cap B}=1$.

By the principle of inclusion--exclusion, we have
$$\abs{A\cup B}=\abs A+\abs B-\abs{A\cap B}=6+6-1=11.$$

\noindent Alternatively, we could have found $\abs{A\cup B}$ by listing all of
the elements of~$A\cup B$ explicitly:
$$\eqalign{
	A\cup B=\{\,&\hbox{1,3,3,3,3,3};\,\hbox{2,3,3,3,3,3};\,
	\hbox{3,3,3,3,3,1};\,\hbox{3,3,3,3,3,2};\,\hbox{3,3,3,3,3,3};\,
	\hbox{3,3,3,3,3,4};\cr
	&\hbox{3,3,3,3,3,5};\,\hbox{3,3,3,3,3,6};\,
	\hbox{4,3,3,3,3,3};\,\hbox{5,3,3,3,3,3};\,\hbox{6,3,3,3,3,3}\,\}.\cr
}$$

\noindent So, using $S$ to denote the sample space, we have
$$\Pr(A\cup B)={\abs{A\cup B}\over\abs S}={11\over46{,}656}.$$
Therefore, the probability that a~``3'' is rolled at least five times in a row
is $11/46{,}656$, or about two~hundredths of one percent. \qed

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem (From {\it The Colossal Book of Short Puzzles and Problems\/} by
Martin Gardner.)\enspace A secretary types four letters to four people and
addresses the four envelopes. If she inserts the letters at random, each in a
different envelope, what is the probability that exactly three letters will go
into the right envelopes?

\solution If three letters go into the right envelopes, then the fourth letter
must go into the fourth envelope, which must be the right one! So the event
that exactly three letters go into the right envelopes contains zero outcomes,
which means that the probability of this event is zero. \qed

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem (From {\it The Colossal Book of Short Puzzles and Problems\/} by
Martin Gardner.)\enspace Bill, a student in mathematics, and his friend John,
an English major, usually spun a coin on the bar to see who would pay for each
round of beer. One evening Bill said: ``Since I've won the last three spins,
let me give you a break on the next one. You spin {\it two\/} pennies and I'll
spin one. If you have more heads than I have, you win. If you don't, I win.''

``Gee, thanks,'' said John.

On previous rounds, when one coin was spun, John's probability of winning was,
of course, $1/2$. What are his chances under the new arrangement?

\solution Let's make a table of the possible outcomes. Remember, John wins if
he gets {\it more\/} heads than Bill. If there is a tie, Bill wins.
$$\vbox{\offinterlineskip
\hrule
\halign{\vrule#&\strut\quad\hfil#\hfil\quad&\vrule#&\strut\quad\hfil#\hfil
	&&\strut\quad\hfil#\hfil\quad&\vrule#\cr
height2pt&\omit&height2pt&\omit&\omit&height2pt&\omit&height2pt\cr
&Bill&&\multispan2\strut\quad\hfil John\hfil\quad&&Winner&\cr
height2pt&\omit&height2pt&\omit&\omit&height2pt&\omit&height2pt\cr
\noalign{\hrule}
height2pt&\omit&height2pt&\omit&\omit&height2pt&\omit&height2pt\cr
&H&&H&H&&John&\cr
&H&&H&T&&Bill&\cr
&H&&T&H&&Bill&\cr
&H&&T&T&&Bill&\cr
&T&&H&H&&John&\cr
&T&&H&T&&John&\cr
&T&&T&H&&John&\cr
&T&&T&T&&Bill&\cr
height2pt&\omit&height2pt&\omit&\omit&height2pt&\omit&height2pt\cr}
\hrule}$$

The eight outcomes in the table above are equally likely. John wins in four of
the eight cases. So the probability that John wins under the new arrangement
is~$4/8$, or~$1/2$, the same as it was before. \qed

\nobreak\medskip

\noindent[Gardner points out that John's probability of winning remains~$1/2$
whenever he has one more coin than Bill. For example, if John has 51~coins and
Bill has~50, John's probability of winning is still~$1/2$.]

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Alice has three fair six-sided dice, colored green, red, and blue, but
these dice are not numbered in the standard way. Instead, they are numbered
like this:
$$\vbox{\halign{\hfil\strut#&&\enspace#\hfil\cr
Green:&5&5&5&2&2&2\cr
Red:&4&4&4&4&4&1\cr
Blue:&3&3&3&3&3&6\cr}}$$
In other words, the green die has three faces labeled~``5'' and three
labeled~``2''; the red die has five faces labeled~``4'' and one labeled~``1'';
and the blue die has five faces labeled~``3'' and one labeled~``6.''

Alice proposes to Bob that they play a little game. ``Here's how it will
work,'' explains Alice. ``In each round, we will each choose one die and
discard the third one. Then we will simultaneously roll our chosen dice.
Whoever rolls the higher number will win a dollar from the other player. (Note
that there can never be a tie.) And since I'm such a nice person, I will let
you choose your die first, before I choose mine.''

\nobreak\smallskip

\item{(a)} Suppose Bob chooses the red die and Alice chooses the green die.
Show that the probability that Alice wins is greater than~$1/2$.

\nobreak\smallskip

\item{(b)} After using the red die for a while and losing more often than not
to Alice's green die, Bob begins to suspect that the odds are not in his favor.
So he chooses the blue die instead, and Alice chooses the red die. Show that
the probability that Alice wins is still greater than~$1/2$.

\nobreak\smallskip

\item{(c)} Bob continues to lose when he plays the blue die against Alice's red
die. He reasons, ``The green die is better than the red die, and the red die is
better than the blue die. So, clearly, I should choose the green die, because
it must be the best of all.'' He takes the green die, and Alice takes the blue
die. Show that the probability that Alice wins is {\it still\/} greater
than~$1/2$.

\nobreak\smallskip

\item{(d)} Alice's dice are an example of {\it nontransitive dice}. How is this
dice game similar to the game of Rock, Paper, Scissors?

\solution

\firstpart{(a)} We draw a probability tree diagram showing Bob's and Alice's
rolls when Bob has the red die and Alice has the green die.
$$\fig{homework-7-sol-red-green-tree.mps}$$
To find the overall probability that Alice wins, we add the probabilities of
each outcome in which Alice wins. So the probability that Alice wins is
$5/12+1/12+1/12=7/12$, which is greater than~$1/2$ (it is about~58\%).

\part{(b)} When Bob has the blue die and Alice has the red die, the probability
tree diagram looks like the following.
$$\fig{homework-7-sol-blue-red-tree.mps}$$
Alice wins with probability~$25/36$, which is greater than~$1/2$ (it is
about~69\%).

\part{(c)} When Bob has the green die and Alice has the blue die, the
probability tree diagram looks like the following.
$$\fig{homework-7-sol-green-blue-tree.mps}$$
Alice wins with probability~$1/12+5/12+1/12=7/12$, which is greater than~$1/2$
(it is about~58\%).

\part{(d)} In the game of Rock, Paper, Scissors, each object wins over one of
the other two objects but loses to the other one: rock beats scissors but loses
to paper, paper beats rock but loses to scissors, and scissors beats paper but
loses to rock. In Alice's dice game, each die ``wins'' over one of the other
two dice (with probability greater than~$1/2$) but ``loses'' to the other one
(with probability greater than~$1/2$): green wins over red but loses to blue,
red wins over blue but loses to green, and blue wins over green but loses to
red. \qed

\nobreak\medskip

\noindent[True story: Warren Buffett once challenged Bill Gates to play this
game, offering to let Gates choose his die first. Gates was intrigued, examined
the dice, figured out the trick, and then demanded that Buffett choose first.]

\nobreak\bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Ten mathematicians are captured by pirates. The pirate captain tells
them, ``I have an unusual custom for dealing with prisoners on my ship.
Tomorrow at dawn I will put all ten of you in a line, blindfolded and facing
the same direction. I will assign each of you either a black hat or a white
hat, depending on the outcome of the flip of a fair coin. (Since there are ten
of you, I will flip the coin ten times, once for each of you.)

``After you have each been given a hat, I will remove your blindfolds. Each of
you will then be able to see the colors of the hats of all the people standing
in front of you in the line, but you will not be able to see your own hat, nor
will you be able to see the hats of the people behind you.

``Then I will begin my questioning. I will start with the person at the back of
the line (who can see all nine of the other hats). I will ask him what color
his hat is. He must announce his guess, either black or white, in a flat,
monotone voice, but loudly enough for all the rest of you to hear. If his guess
is correct, I will allow him to go free. If his guess is incorrect, however, he
will be killed.

``Next, I will move to the second-to-last person in line (who can see eight
other hats), and I will repeat the procedure, moving up the line, until I have
dealt with each of you in turn. By the way, once the hats have been
distributed, you will not be allowed to pass any information among yourselves
except your guesses.

``Well, I suppose it's getting pretty late, and we have a big day tomorrow.
Sleep well, and I'll see you in the morning!''

That night the mathematicians get together and decide that they must devise a
strategy to maximize the expected number of them who will go free. What
strategy should they adopt?

\solution The surprising answer is that nine of the ten mathematicians can be
saved with certainty!

Let's number the mathematicians 1 through~10, where mathematician~1 is at the
front of the line (and can therefore see no hats) and mathematician~10 is at
the back of the line (and can see all nine of the other hats).

The key to the best strategy is the following: Mathematician~10, who is the
first to guess, counts the number of white hats he sees in front of him. If he
sees an even number of white hats, he guesses ``white''; if he sees an odd
number of white hats, he guesses ``black.'' Since all of the remaining
mathematicians can hear his guess, they will know whether there is an even
number or an odd number of white hats among them.

When it is time for mathematician~9 to guess, he can see the number of white
hats in front of him, and he knows (from mathematician~10's guess) whether
there is an even number or an odd number of white hats among mathematicians 1
through~9. This information is enough to allow mathematician~9 to deduce the
color of his hat with certainty. For instance, if he sees an even number of
white hats in front of him and he knows that mathematician~10 also saw an even
number of white hats, then his own hat must be black. On the other hand, if he
sees an even number of white hats in front of him and mathematician~10 saw an
{\it odd\/} number of white hats, then his own hat must be white. So
mathematician~9 can be assured of guessing his own hat color correctly.

Mathematician~8 can use similar reasoning to deduce the color of his hat. When
it is his turn to guess, he can see the seven hats in front of him, he knows
whether mathematician~10 saw an even number or an odd number of white hats, and
he also knows the color of mathematician~9's hat (from mathematician~9's guess,
which was correct). This is enough information to allow him to determine his
own hat color, so he is also assured of guessing his own hat color correctly.

In general, when it is time for mathematician in the middle of the line to
guess, the information available to him includes the colors of all the hats in
front of him (which he can see directly); whether mathematician~10 saw an even
number or an odd number of white hats (from mathematician~10's guess); and the
colors of all the hats behind him, except the hat of mathematician~10 (from the
guesses of the other mathematicians). From this information he can work out
what color his own hat must be, and so he will be able to guess correctly
(simultaneously passing the information about the color of his own hat to the
people in front of him, who will need it to determine the colors of their
hats).

Therefore, if the mathematicians follow this strategy, mathematicians 1
through~9 will be saved with certainty. Unfortunately, mathematician~10 will
only guess correctly half the time (by sheer luck), but, as he has no possible
source of information about the color of his own hat, this is the best that can
be done. In other words, with probability~$1/2$ it will be the case that nine
mathematicians go free, and with probability~$1/2$ it will be the case that all
ten mathematicians go free. The expected number of survivors is therefore
$$\frac12(9)+\frac12(10)=9.5,$$
which is really quite impressive. \qed

\bigbreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Suppose the disease diauropunctosis afflicts 1 out of every 5,000
people in the United States. (Assume that the occurrence of the disease is
completely random, so that each person has a 1/5,000 probability of having the
disease, independently of everyone else.) Fortunately, there is a test for
diauropunctosis, which is 99\%~accurate, meaning that 1\% of the people who are
tested receive incorrect results. (Assume that the accuracy of the test is
independent of the occurrence of diauropunctosis; in other words, assume the
test is 99\%~accurate for people who have the disease and 99\%~accurate for
people who do not have the disease.) The city of Pierce, population~1,000,000,
has received a government grant to test all of its citizens for
diauropunctosis.

\nobreak\smallskip

\item{(a)} About how many people in Pierce have diauropunctosis?

\nobreak\smallskip

\item{(b)} Out of the people who have diauropunctosis, about how many will test
positive for the disease? About how many will test negative?

\nobreak\smallskip

\item{(c)} Out of the people who do {\it not\/} have diauropunctosis, about how
many will test positive? About how many will test negative?

\nobreak\smallskip

\item{(d)} Out of all the people who test positive for diauropunctosis, what
percentage actually have the disease?

\nobreak\smallskip

\item{(e)} Think about what your answer to part~(d) means. Suppose you are
tested for diauropunctosis, and the test results come back positive. What is
the (conditional) probability that you actually have diauropunctosis, given
that you tested positive? Explain why receiving a positive result from a test
that is 99\%~accurate does not mean that you have a 99\%~probability of having
the disease.

\solution

\def\0{\phantom0}

\firstpart{(a)} Since the prevalence of diauropunctosis is 1~in~5,000, the
number of people in Pierce who have diauropunctosis is about
$${1\over5{,}000}(1{,}000{,}000)=200.$$

\part{(b)} The test for diauropunctosis is 99\%~accurate, which means that 99\%
of the people who are tested will receive the correct result (and the remaining
1\% will receive the wrong result). Therefore, out of the 200~people in Pierce
who have diauropunctosis, about
$$\eqalign{
	0.99(200)&=\hbox{198 will test {\it positive\/};}\cr
	0.01(200)&=\hbox{\0\02 will test {\it negative}.}\cr
}$$

\part{(c)} Out of the 999,800~people in Pierce who do not have diauropunctosis,
about
$$\eqalign{
	0.99(999{,}800)&=\hbox{989,802 will test {\it negative\/};}\cr
	0.01(999{,}800)&=\hbox{\0\09,998 will test {\it positive}.}\cr
}$$

\part{(d)} There will be about 198~people who test positive and actually have
diauropunctosis, and about 9,998~people who test positive but do not have the
disease. So in all there will be about 10,196~people who test positive. Out of
these, the percentage who actually have the disease is about
$${198\over10{,}196}\approx0.0194=1.94\%.$$

\part{(e)} The answer to part~(d) means that the conditional probability that a
person has diauropunctosis, given that the person tested positive, is
about~1.94\%. The positive test result significantly increased the probability
that the person has the disease (from the original probability of~$1/5{,}000$
to about~$1/51$), and so follow-up tests are appropriate, but because of the
rarity of the disease it is far more likely that the positive test result was
in error than that the person actually has diauropunctosis. The simple truth is
that healthy people vastly outnumber people with the disease, and so false
positives (though relatively rare with such an accurate test) will still be
much more common than actual occurrences of the disease. \qed

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

\bye
