\problem {\bf Pharmaceutic Food} 

Farmer John's cows occasionally suffer from the famous "Mad Cow" disease.
Fortunately, they recover from it fairly nicely, although it may take
up to four weeks.  Half of the cows that contract the disease recover
in two weeks, and the other cows all recover in four weeks.  Farmer
John is trying a new treatment for the disease, a special feed for those
cows that have the disease.  For cows that respond to this treatment, all
of them recover in two weeks.  Unfortunately, only 70\% of the cows
respond to the treatment.

Farmer John's prize cow, Elsie, contact the disease.  Farmer John, of
course, puts Elsie on the special feed, and Elsie recovers in two weeks.
What is the probability that Elsie is responsive to the medicine?

\problem
In London, half of the days have some rain. The weather forecaster is
correct 2/3 of the time -- i.e. the probability that it rains, given
that she has predicted that it will rain, and the probability that it
does not rain given that she has predicted that it does not rain, are
both equal to 2/3. When rain is forecast, Mr. Pickwick takes his
umbrella. When rain is not forecast, he takes it with probability
1/3. Find: \\
a) the probability that Pickwick has no umbrella, given that it rains. \\
\solution{\\
We construct a tree as in lecture notes \#19 (see figure 3). For part (a), we wi
sh to find
\prob{\text{no umbrella} \mid \text{rain}}
\[ = \frac{\prob{\mbox{no umbrella} \cap \mbox{rain}}}
          {\prob{\mbox{rain}}} \]
\[ = \frac{1/9}{1/2} = 2/9 \]
}
b) the probability that it doesn't rain, given that he brings his umbrella.  \\
\solution{
\[ \mbox{\prob{\text{no rain} \mid \text{umbrella}}} = \frac{\prob{\mbox{no rain} \cap \mbox{u
mbrella}}}
          {\prob{\mbox{umbrella}}} \]
\[ = \frac{1/6 + 1/9}{1/9 + 1/6 + 1/3 + 1/18} = 5/12 \]
}



\problem \textbf{Sample Space. Events. Conbinations of Probabilites.}

\bparts

 ``Concrete Mathematics.'' Graham, Knuth, Patashnik. 
\ppart We are rolling a pair of dice.  Describe the sample set of
elementary events. 

\solution{The set $\Omega$ of elementary events is $D^2 = {(1,1),
(1,2),\ldots,(6,6)}$, where $D = {1, 2, 3, 4, 5, 6}$ is the set of all
six ways that a given die can land.  Two rolls such as $(1,2)$ and
$(2,1)$ are considered to be distinct; hence this probability space
has a total $6^2 = 36$ elements.}

\ppart Let $\mathcal{A}$ be the event that doubles are thrown. What's
the probability of $\mathcal{A}$?

\solution{$\prob{\omega \in \mathcal{A}} = \sum_{w\in
\mathcal{A}}\prob{\omega}$.  Thus, the probability of doubles with fair
dice is $\frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} +
\frac{1}{36} + \frac{1}{36} = \frac{1}{6}$}

\ppart Now suppose that not all dice are ``fair.'' Let $\prob{1} = \prob{6}
= \frac{1}{4}; \prob{2} = \prob{3} = \prob{4} = \prob{5} = \frac{1}{8}$. What's
$\prob{\omega \in \mathcal{A}}?$

\solution{$\prob{\omega \in \mathcal{A}} = 4 \times \frac{1}{8} + 2 \times \frac{1}{4} = \frac{3}{16}$}

%http://www.unc.edu/~rowlett/Math81/problem_sets/Set11.html
\ppart Going back to fair dice, which is more likely, rolling 8 with
two dice or rolling 8 with three dice?

\solution{If we roll two dice, there are 36 possible outcomes, of
which 5 (6-2, 5-3, 4-4, 3-5, and 2-6) yield an 8. So the probability
of rolling 8 with two dice is $\frac{5}{36} = 0.1389$. If we roll
three dice, there are 63 = 216 possible outcomes. Exactly 21 of these
outcomes yield 8 (you can list them), so the probability of rolling 8
with three dice is $\frac{21}{216} = 0.0972$. Thus it is more likely
to roll 8 with two dice.}

\eparts 
\problem

Suppose that a resident of Pittsburgh chosen at random is
a Steelers fan with probability $0.90$, that a resident chosen at
random is a Penguins fan with probability $0.75$, and that a resident
Steelers fan chosen at random is a Penguins fan with probability
$0.80$.  Now suppose that a resident is chosen at random from among
those that are {\it not\/} Steelers fans.  What is the probability
that such a resident is a Penguins fan?

Let $P$ represent the event that a resident is a Penguins fan, and $S$ represent the event
that the resident is a Steelers fan.  Then:
 
\begin{eqnarray*}
\prob{P} & = & \prob{P \mid  S} \prob{S} + \prob{P \mid  \overline{S}} \prob{\overline{S}} \\
\prob{P \mid  \overline{S}} & = & \frac{\prob{P} - \prob{P \mid  S} * \prob{S}}{\prob{\overline{S}}} \\
& = & \frac{0.75 - 0.80 \cdot 0.90}{1 - 0.9} \\
& = & 0.3
\end{eqnarray*}

\problem \textbf{Conditional Probability}
% 18.313 Quiz I, problem 4. 

Suppose that 5 men out of 100, and 25 women out of 10,000 are
colorblind.  A colorblind person is chosen at random from a population
with equal numbers of males and females.  What is the probability of
this person being male?

\solution{Let M be the event the person is male, F female,
C colorblind.  We are given $\prob{C \mid M} = \frac{5}{100}$ and
$\prob{C \mid F}=\frac{25}{10,000}.$ Since the population consists of equal
numbers of males and females, $\prob{M} = \prob{F} = \frac{1}{2}$.  So,
\[\prob{M \mid C} = \frac{\prob{C \mid M}\times \prob{M}}{\prob{C}} = \frac{\prob{C \mid M}\times \prob{M}}{\prob{C \mid M}\times \prob{M} + \prob{C \mid F}\times \prob{F}} = \]
\[\frac{(5/100)\times (1/2)}{(5/100)\times (1/2) + (25/10000) \times
(1/2)} = 20/21. \]}

\problem

Suppose that a resident of Pittsburgh chosen at random is
a Steelers fan with probability $0.90$, that a resident chosen at
random is a Penguins fan with probability $0.75$, and that a resident
Steelers fan chosen at random is a Penguins fan with probability
$0.80$.  Now suppose that a resident is chosen at random from among
those that are {\it not\/} Steelers fans.  What is the probability
that such a resident is a Penguins fan?

Let $P$ represent the event that a resident is a Penguins fan, and $S$ represent the event
that the resident is a Steelers fan.  Then:
 
\begin{eqnarray*}
\prob{P} & = & \prob{P  \mid  S} \prob{S} + \prob{P  \mid  \overline{S}} \prob{\overline{S}} \\
\prob{P  \mid  \overline{S}} & = & \frac{\prob{P} - \prob{P  \mid  S} * \prob{S}}{\prob{\overline{S}}} \\
& = & \frac{0.75 - 0.80 \cdot 0.90}{1 - 0.9} \\
& = & 0.3
\end{eqnarray*}
\problem 
% 18.313 Quiz I, problem 3

A population consists of $c_1$ members of type 1, $c_2$ members
of type $2,\cdots ,c_n$ members of type $n$.  A set of $k$ members is
drawn from the population without replacement. 

\bparts

\ppart Find the probability that $i$ members of type 1
were chosen.

\solution{
Let $N_{1i}$ be the event that $i$ members of class 1 were
chosen.  $\prob{N_{1i}} = \frac{\binom{c_1}{i} \times \binom{c_2+\cdots +
c_n}{k-i}}{\binom{c_1+\cdots +c_n}{k}}$.

}

\ppart Find the probability that $i$ members of class 1
and $j$ members of class 2 were chosen.

\solution{Let $N_{2j}$ be the event that $j$ members of class 2 were
chosen.  Then $ \prob{N_{1i} \cap N_{2j}} = \frac{\binom{c_1}{i} \binom{c_2}{j}
\binom{c_3 +\cdots + c_n}{k-i-j}}{\binom{c_1 + \cdots +c_n}{k}} $}

\eparts

\problem
In London, half of the days have some rain. The weather forecaster is
correct 2/3 of the time -- i.e. the probability that it rains, given
that she has predicted that it will rain, and the probability that it
does not rain given that she has predicted that it does not rain, are
both equal to 2/3. When rain is forecast, Mr. Pickwick takes his
umbrella. When rain is not forecast, he takes it with probability
1/3. Find: \\
a) the probability that Pickwick has no umbrella, given that it rains. \\
b) the probability that it doesn't rain, given that he brings his umbrella.  \\
\solution{\\
We construct a tree as in lecture notes \#19. For part (a), we wish to find 
\begin{eqnarray*}
\prob{\text{no umbrella}  \mid \text{rain}}
  & = &\frac{\prob{\mbox{no umbrella} \wedge \mbox{rain}}}{\prob{\mbox{rain}}}\\
& = & \frac{1/9}{1/2} = 2/9.
\end{eqnarray*}

For part (b), we wish to find
\begin{eqnarray*}
\prob{\text{no rain} \mid \text{umbrella}} & = &
 \frac{\prob{\mbox{no rain} \wedge \mbox{umbrella}}}{\prob{\mbox{umbrella}}}\\
& = & \frac{1/6 + 1/9}{1/9 + 1/6 + 1/3 + 1/18} = 5/12
\end{eqnarray*}

}

\problem
{\bf Three-Cornered Duel.} This problem involves old-fashioned duels
by pistol.  [You may want to use a calculator for this problem.]

\bparts
\ppart
A and C are engaged in a duel.  Each takes turns firing
a shot at the other until one of them is hit and falls down. On each
shot, A has probability 0.3 of hitting, and C has probability 0.5. If
A shoots first, what is the chance that A will win this duel?

\solution{
One approach we might take to this problem uses infinite sums, as in
the craps problem above.  In other words, suppose $p_i$ is the
probability that A wins after $i$th turns.  To win on the first turn,
A must hit immediately, so $p_0 = 0.3$.  To win on the second turn, A
must miss, then B must miss, then A must hit, giving $p_1 = 0.7 \cdot
0.5 \cdot 0.3$.  
In general, $p_i = (0.7 \cdot 0.5)^i 0.3$.  
Summing this over all $i$ gives:

\begin{eqnarray*}
\sum_{i=0}^\infty p_i & = & \sum_{i=0}^\infty (0.7 \cdot 0.5)^i 0.3 \\
& = & 0.3 \cdot \frac{1}{1 - {0.7 \cdot 0.5}} \\
& \approx & 0.462
\end{eqnarray*}

Another approach relies on the problem's {\it self-similarity.}
Suppose A misses on the first turn, then B misses.  Now the situation
is exactly the same as it was at the beginning: it's A's turn, and
both duellists still have the same probability of hitting.  So the
probability that A wins given that A just missed and B just missed
must be the same as the overall probability that A wins.  If we let
$p$ be the probability that A wins overall, we can write:

\begin{eqnarray*}
\prob{\mbox{A wins eventually}} & = & \prob{\mbox{A hits on first turn}} + \prob{\mbox{A hits after the first turn}} \\
\prob{\mbox{A wins eventually}} & = & \prob{\mbox{A hits on first turn}} \\
& & + \prob{\mbox{A wins eventually}  \mid  \mbox{A missed then B missed}}\prob{\mbox{A missed then B missed}} \\
p & = & 0.3 + 0.7 \cdot 0.5 p \\
(1 - 0.35)p & = & 0.3 \\
p & \approx & 0.462
\end{eqnarray*}

}
\ppart A, B, and C are to fight a ``three-cornered duel.''
They take turns (in the order A,B,C, cyclically) firing at their
choice of target.  Once hit, a duellist falls down, dropping out of
the duel. The duel continues until only one is left standing. They all
know that A has probability 0.3 of hitting a target, and C has
probability 0.5. But B never misses! (Assume that B is rational, and
always eliminates the most serious threat.)  What should A's strategy
be?

\solution{
If B has a chance, B will shoot and kill C in the first round, because
C is the most serious threat.  So A must decide whether to shoot at B
(which might be wise, because B is incredibly dangerous) or at C
(which might be better, because leaving B alive ensures that C won't
get a shot at either of them).

Suppose A aims at B in
the first round.  If A hits B (probability 0.3), and C shoots at and
misses A (probability 0.5), then A and C will be left in a duel
identical to the warmup problem (with probability 0.462 that A wins).
If A misses B (probability 0.7), then B will hit C (probability 1),
and A will get one more chance to hit B before certain annihilation
(probability 0.3).  So the probability that A wins by aiming at B is about
$0.3 \cdot 0.5 \cdot 0.462 + 0.7 \cdot 0.3 = 0.279$.

Now suppose A aims at C in the first round.  If A hits C, then A
definitely loses, since B will unerringly shoot and hit A.  If A
misses C (probability 0.7), then B will take out C and A will get one
chance at B (probability 0.3).  So the probability that A wins by
aiming at C is exactly $0.7 \cdot 0.3 = 0.21$.

But there's a third alternative: A can intentionally miss everybody.
Then B shoots C, and A has one chance to hit B (probability 0.3).  So
the probability that A wins by shooting in the air is exactly $0.3$,
which makes it the best strategy.}
\eparts
\fi
