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%spring98 final #7 (added by Ivan) %formerly F97 9 pfinal
\problem\points{20}

A high blood pressure test is known to be $90\%$ accurate in detecting
high blood pressure (that is, $10\%$ false negatives), whereas only $5\%$
of people without high blood pressure will falsely test positive.  Suppose
$30\%$ of the population has high blood pressure.  Given that a person
picked at random tests positive, what is the conditional probability that
this person has high blood pressure?

\medskip
\solution{
\textbf{Solution:}
Let $T = \{\text{test is positive} \}$, $V = \{\text{person has HBP}\}$.
We are given that $P(T|V) = 0.9$, $P(T|\bar{V}) = 0.05$ and $P(V) = 0.3$.

\[
\pr{V|T} = \frac{P(T|V)P(V)}{P(T)} =
\frac{P(T|V)P(V)}{P(T|V)P(V)+P(T|\bar{V})P(\bar{V})}
\]

The above expression evaluates to $0.885$.
}

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%Fall 1998 Final exam
% Added by George
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\problem {\bf (5 points)} A spider eats three flies a day.  Each fly
attempting to pass through his web has an even chance of becoming
entangled.  The spider eats the flies that become entangled in his web
until he fills his quota.  After his quota is filled, the spider
allows flies that become entangled to work themselves free and escape.

Wilbur the Fly is about to attempt a flight through the spider's web.
If five flies have already made the attempt today, then what is the
probability that Wilbur survives?

(Assume that the flies become entangled mutually independently.)

\solution{Among the first five flies, the number that become entangled
is either less than half ($\leq 2$) or more than half ($\geq 3$).
These two events are equally likely, since each fly has an even chance
of becoming entangled.  Therefore, the spider has an even chance of
filling his quota with the first five flies.

The probability that Wilbur survives is computed below.

\begin{eqnarray*}
\Pr(\text{fly lives})
    & = &   1 - \Pr(\text{Wilbur dies}) \\
    & = &   1 - \Pr(\text{quota unfilled} \cap \text{Wilbur entangled}) \\
    & = &   1 - \Pr(\text{quota unfilled}) \cdot
            \Pr(\text{Wilbur entangled}) \\
    & = &   1 - \frac{1}{2} \cdot \frac{1}{2} \\
    & = &   \frac{3}{4}
\end{eqnarray*}

In the first step, we switch to computing the probability of the
complementary event.  The second step uses the fact that Wilbur the
Fly dies only if the spider's quota is unfilled and the fly becomes
entangled.  The third step follows from the fact that flies become
entangled mutually independently.  The fourth step is substitution, and
the last step is simplification.}

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%Fall 2000 Final exam
% Added by George
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%% PROBLEM 11
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\problem\pts{10} \textbf {Chess Matches}

Math Moose and Theory Pig have been playing a series of chess games,
and have noticed a number of facts:

\begin{itemize}
\item  Math Moose plays white in any given game with probability 8/17.

\item If Math Moose plays white, he is equally likely to win or lose.

\item If Math Moose plays black, then with probability 8/9, he loses.

\item If Math Moose wins, he always takes Theory Pig's queen at
some point in the game.

\item If Math Moose plays white and loses, he takes Theory Pig's queen
with probability 1/2.

\item If Math Moose plays black and loses, he takes Theory Pig's queen
with probability 1/4.
\end{itemize}

\problempart {\bf (4 points)} Define the sample space, and give the
probability associated with each outcome.

\solution{\bigskip {\bf Solution:}
The outcomes are that Math Moose...:

plays white, wins, and takes the queen: 4/17\\
plays white, loses, and takes the queen: 2/17\\
plays white, loses, and doesn't take the queen: 2/17\\
plays black, wins, and takes the queen: 1/17\\
plays black, loses, and takes the queen: 2/17\\
plays black, loses, and doesn't take the queen: 6/17\\
plays white, wins, and doesn't take the queen: 0\\
plays black, wins, and doesn't take the queen: 0\\
}
\smallspace
\smallspace


\problempart {\bf (3 points)} What is the probability that Math Moose
plays white, given that he wins?

\solution{\bigskip {\bf Solution:}
The probability is $\frac{4/17}{5/17} = \frac{4}{5}$.}
\smallspace


\problempart {\bf (3 points)} What is the probability that Math Moose
wins, given that he takes Theory Pig's queen?

\solution{\bigskip {\bf Solution:}
The probability is $\frac{5/17}{9/17} = \frac{5}{9}$.}
