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\def\header{{\hfill Exam 2
\hfill Canary Yellow
\hfill A. Miller
\hfill Fall 2005
\hfill Math 210\hfill}}

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\begin{document}

\bigskip



{ \bf Show all work.

Simplify your answer.

Circle your answer. }

\bigskip

No books, no calculators, no cell phones, no pagers, no
electronic devices of any kind.


\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\bigskip
Circle your Discussion Section:
\begin{verbatim}

   343  T   12:05--12:55        1412 STERLING
   344  R   12:05--12:55        1327 STERLING
   345  T   13:20--14:10        1327 STERLING
   346  R   13:20--14:10        1327 STERLING 

\end{verbatim}
\bigskip

\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
   \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}

\immediate\openout\examTBLFile=exam.tbl

\bigskip

Solutions will be posted shortly after the exam:
www.math.wisc.edu/$\sim$miller/m210

\setcounter{page}{0}

\newpage

\prob{10} % 3.1
State the three Axioms for a Probability Measure:

\bigskip\bigskip\bigskip\bigskip
A probability measure assigns to each event $E$ of a sample space $S$
a number denoted by $Pr[E]$ and called the probability of $E$. This
assignment must satisfy the three axioms:

\bigskip\bigskip\bigskip\bigskip
i.

\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip
ii.

\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip
iii. 

\answer see page 88.

\prob{10} % 3.R-15 on 3.2
There are 9 mice in a cage: 3 white males, 4 gray females, and 2 gray males.
Two mice are selected simultaneously and at random. Find the probability
that at least one mouse is a male, given that at least one is gray.

\answer $9/11$


\prob{10} % 3.3-23
There are 5 quarters, 1 dime, and 3 nickels in a drawer. An experiment
consists of selecting a coin at random, noting its value, and setting
it aside. If it is a dime, the experiment ends. If it is not a dime,
then another coin is selected at random, and its value noted. 
Find the probability that at least one nickel is selected.

\answer $13/24$.  This assumes that at most two selections were made which was
what was intended.  \par Some students interpreted the problem to mean that the
experiment continues until a dime is selected. We may as well assume that the
selection process stops whenever a nickel or dime is selected.  Since any of
the four stopping coins is equally likely to be the stop coin, the probability
that the stop coin is a nickel is $3/4$.  This probability is the same whether
there is 0 quarters or 100 quarters.


\prob{10} % 3.4-18
Students are being tested for Virus X and it is estimated that
one percent of the students are infected.  If a student is infected, the test
is positive 80\% of the time.  If a student is not infected, the test
is negative 90\% of the time. If the test is applied to a student whose
infection status is unknown, and if the test is negative, find the 
probability that the student is actually infected with Virus X.

\answer $2/893$

\prob{10} % 3.5-23
A high school basketball player makes one-third of his three-point shots.
If we assume that his shots are Bernoulli trails, how many must he shoot
to have a probability of at least $3/4$ of making at least one of them?

\answer $n=4$ this is the least $n$ such that $(2/3)^n\leq 1/4$.



\prob{10} % 4.R- 7
A carnival game consists of selecting 3 balls simultaneously and at random from
a box containing 3 red and 5 green balls. Each red ball pays $50$ cents each
green ball pays 10 cents.  It costs 1 dollar to play.  A random variable $X$ is
the net payoff, i.e., prize money minus cost. 
Find the probability density function of $X$.

\answer \par\bigskip \begin{tabular}{c|c}
k & Pr(X=k) \\
\hline    
.5 & 1/56 \\
.1 & 15/56 \\
-.3 & 30/56 \\\
-.7 & 10/56 \\
\end{tabular}


\prob{10} % 4.2-13
A coin is weighted so that the probability of
{\bf Heads} is $2/5$.  The coin is flipped ten
times. Let $X$ be the random variable which counts the
number of {\bf Tails} which come up.  Find the 

\bigskip\bigskip

(a) expectation of $X$,
$\mu=E(X)$ 


\bigskip\bigskip
$\mu=$

\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip
(b) the variance of $X$, $\nu=Var(X)$, 

\bigskip\bigskip
$\nu=$

\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip

\bigskip\bigskip
(c) and the standard
deviation of $X$, $\sigma=SD(X)$.

\bigskip\bigskip
$\sigma=$

\bigskip\bigskip\bigskip\bigskip\bigskip\bigskip

\answer $\mu=6$, $\nu=12/5$, $\sigma=\sqrt{\nu}$


\finishtbl

\printanswers

\end{document}




