\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{verbatim}
\allowdisplaybreaks

\pagestyle{plain} \setlength{\topmargin}{-0.5in}
\setlength{\textheight}{9.85in} \setlength{\oddsidemargin}{-.1in}
\setlength{\evensidemargin}{-.2in} \setlength{\textwidth}{6.8in}
\font\heada=cmbx10 scaled\magstep3 \font\headb=cmsl10
scaled\magstep1 \font\headc=cmr8 \pretolerance=10000

\begin{document}


\section*{Chapter 6 Problems from the textbook}
\subsection*{Exercises} 6.1 on p258: 1,3, 5, 9

\noindent 6.2 on p263: 15, 17

\noindent 6.3 on p273: 19, 21

\subsection*{Reading} Sections 6.1-6.3

\section*{Key to selected problems} The answers in the back of the
textbook for problem 9 on page 258 and problem 21 on page 273 are
incorrect.   The correct answers are:

\noindent 6.9 on page 258:
\begin{enumerate}
\item[a] Assuming that Jeanie forgetting any errand is independent
of whether she forgets another one, then the probability that she
forgets ALL three errands is $(.1)^3=.001$

\item[b] The event that Jeanie remembers at least one errand is
the same as the probability that she does not forget all of the
errands: $1 - (.1)^3 = .999$

\item[c] $(.9)(.1)^2=.009$
\end{enumerate}

\noindent 6.21 on page 273
\begin{enumerate}
\item[a] $P({\rm male})=\frac{200 + 3200 + 2500 + 1500 + 2100 +
1500+200}{18000}=\frac{11200}{18000}= .6\bar 2$

\item[b] $P({\rm College~of~
Ag})=\frac{2100+900}{18000}=\frac{3000}{18000}= .1\bar 6$

\item[c] $P({\rm male}|{\rm Ag})=\frac{2100}{3000}=.7$.   Thus,
the probability $P({\rm male ~\underline{and}~ Ag}) = P({\rm
male}|{\rm Ag})P({\rm Ag}) = .7(.1\bar 6)=.11\bar 6$.   Or,
directly from the table, we see that $P({\rm male
~\underline{and}~ Ag})=\frac{2100}{18000}=.11\bar 6$

\item[d] $P({\rm
male~}|\rm{~not~in~Ag})=\frac{11200-2100}{18000-3000}=.60\bar 6$.
So  $P({\rm male ~\underline{and}~ not~in~Ag}) = P({\rm
male~}|\rm{~not~in~Ag})P(\rm{~not~in~Ag})=P({\rm
male~}|\rm{~not~in~Ag})(1-P(\rm{Ag}))=.60\bar 6(1-.1\bar
6)=.50\bar 5$.   Or, directly from the table, $P({\rm male
~\underline{and}~ not~in~Ag})=\frac{9100}{18000}=.50\bar 5$
\end{enumerate}




\end{document}