\documentclass[a4paper,12pt]{article}
\setlength{\textwidth}{6.5in} \setlength{\textheight}{9in}
\setlength{\topmargin}{-.25in} \setlength{\oddsidemargin}{-.25in}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION

An oil company is considering an offshore drilling venture. A
preliminary survey indicates that there may be large ($L$) or
small ($S$) deposits with probabilities ${\rm P}(L) = 0.1$ and
${\rm P}(S) = 0.9$.  Exploration costs \pounds$1\,000\,000$.  If
large deposits are found, the company makes a profit of
\pounds$16\,000\,000$, less the exploration cost.  If small
deposits are discovered, the project will be abandoned.

A team of experts can be employed for \pounds$500\,000$ to perform
a survey.  They would find that either large deposits are likely
($P$), large deposits are unlikely ($N$), or there is inconclusive
evidence ($I$).  The previous record of experts is shown in the
following table.

\begin{center}
\begin{tabular}{cccc}
\hline &\multicolumn{3}{l} {Prediction}\\ \cline{2-4} Actual
result &$P$&$N$&$I$\\ \hline L&60\%&20\%&20\%\\ S&10\%&80\%&10\%
\\ \hline
\end{tabular}
\end{center}

\begin{description}
\item[(a)] Determine the probabilities of the predictions ${\rm P}(P)$,
${\rm P}(N)$ and ${\rm P}(I)$.
\item[(b)] Determine the posterior probabilities for each possible
prediction.
\item[(c)] Use a decision tree to find what course of action the company
should follow.
\end{description}

\bigskip

ANSWER


The table gives

\begin{tabular}{ccc}
P(P|L)=0.6&P(N|L)=0.2&P(I|L)=0.2\\
P(P|S)=0.1&P(N|S)=0.8&P(I|S)=0.1
\end{tabular}

\begin{description}

\item[(a)]

\begin{eqnarray*}
P(P)&=&P(P|L)P(L)+P(P|S)P(S)=(0.6\times0.1)+(0.1\times0.9)=0.15\\
P(N)&=&P(N|L)P(L)+P(N|S)P(S)=(0.2\times0.1)+(0.8\times0.9)=0.74\\
P(I)&=&P(I|L)P(L)+P(I|S)P(S)=(0.2\times0.1)+(0.1\times0.9)=0.11
\end{eqnarray*}

\item[(b)]

\begin{eqnarray*}
P(L|P)&=&\frac{P(P\cap L)}{P(P)}=\frac{P(P|L)P(L)}{P(P)}=0.4\\
P(S|P)&=&\frac{P(P\cap S)}{P(P)}=\frac{P(P|S)P(S)}{P(P)}=0.6\\
P(L|N)&=&\frac{P(N\cap L)}{P(N)}+\frac{P(N|L)P(L)}{P(N)}=0.027\\
P(S|N)&=&\frac{P(N\cap S)}{P(N)}=\frac{P(N|S)P(S)}{P(N)}=0.973\\
P(L|I)&+&\frac{P(I\cap L)}{P(I)}=\frac{P(I|L)P(L)}{P(I)}=0.182\\
P(S|I)&=&\frac{P(I\cap S)}{P(I)}=\frac{P(I|S)P(S)}{P(I)}=0.818
\end{eqnarray*}

Returns (without survey) are:

\begin{tabular}{c|cc}
&Action\\ &$E$&$\overline{E}$\\ \hline L&15&0\\ S&-1&0
\end{tabular}
$E$=explore, $\overline{E}$=do not explore


\setlength{\unitlength}{0.4in}

\begin{picture}(10,20)
\put(10,20){14.5} \put(10,19){$-1.5$} \put(10,18){$-0.5$}
\put(10,17){$-0.5$} \put(10.2,16){14.5} \put(10,15){$-1.5$}
\put(10,14){$-0.5$} \put(10,13){$-0.5$} \put(10,12){14.5}
\put(10,11){$-1.5$} \put(10,10){$-0.5$} \put(10,9){$-0.5$}

\put(0,9){\framebox(.3,.3)} \put(3.3,14.2){\circle{.3}}
\put(3.3,3.8){\framebox(.3,.3)}

\put(.3,9.3){\line(3,5){3}} \put(.3,9){\line(3,-5){3}}

\put(6,17.9){\framebox(.3,.3)} \put(5.7,18.3){4.9}
\put(6,14.2){\framebox(.3,.3)} \put(5.7,14.7){$-0.5$}
\put(6,10.5){\framebox(.3,.3)} \put(5.7,11){1.412}
\put(6,5.8){\circle{.3}} \put(5.7,6.2){0.6} \put(6,2){\circle{.3}}
\put(5.7,2.4){0}

\put(3.6,14.2){\line(3,4){2.4}} \put(3.6,14.2){\line(1,0){2.4}}
\put(3.6,14.2){\line(3,-4){2.4}} \put(3.6,4.1){\line(3,2){2.4}}
\put(3.6,3.8){\line(3,-2){2.4}}

\put(8,19.5){\circle{.3}} \put(7.7,19.8){4.9}
\put(8,17.5){\circle{.3}} \put(7.7,17.8){$-0.5$}
\put(8,15.5){\circle{.3}} \put(7.5,15.8){$-1.068$}
\put(8,13.5){\circle{.3}} \put(7.7,13.8){$-0.5$}
\put(8,11.5){\circle{.3}} \put(7.7,11.8){1.412}
\put(8,9.5){\circle{.3}} \put(7.7,9.8){$-0.5$}

\put(8,6.9){15} \put(8,4.9){$-1$} \put(8,2.9){0} \put(8,.9){2.5}

\put(2,13){\line(1,-1){1}} \put(1,12){survey} \put(1,6){no survey}

\put(3.3,14.8){0.52} \put(3.3,4.5){0.6}

\put(8.2,19.5){\line(4,1){1.8}} \put(8.2,17.5){\line(4,1){1.8}}
\put(8.2,15.5){\line(4,1){1.8}} \put(8.2,13.5){\line(4,1){1.8}}
\put(8.2,11.7){\line(4,1){1.8}} \put(8.2,9.5){\line(4,1){1.8}}

\put(8.2,19.5){\line(4,-1){1.8}} \put(8.2,17.5){\line(4,-1){1.8}}
\put(8.2,15.5){\line(4,-1){1.8}} \put(8.2,13.5){\line(4,-1){1.8}}
\put(8.2,11.7){\line(4,-1){1.8}} \put(8.2,9.5){\line(4,-1){1.8}}

\put(6.3,18.2){\line(4,3){1.6}} \put(6.3,17.9){\line(2,-1){1.6}}
\put(6.3,14.5){\line(4,3){1.6}} \put(6.3,14.2){\line(2,-1){1.6}}
\put(6.3,10.8){\line(4,3){1.6}} \put(6.3,10.5){\line(2,-1){1.6}}
\put(6.3,5.7){\line(4,3){1.6}} \put(6.3,5.7){\line(2,-1){1.6}}
\put(6.3,1.9){\line(4,3){1.6}} \put(6.3,1.9){\line(2,-1){1.6}}

\put(4.3,16.8){P 0.15} \put(4.3,14.4){N 0.74} \put(4.3,11.7){I
0.11}

\put(8.8,20){L 0.4 } \put(8.8,18.7){S 0.6 } \put(8.8,18){L}
\put(8.8,16.7){S} \put(8.8,16){L 0.027 } \put(8.8,14.7){S 0.973 }
\put(8.8,14){L} \put(8.8,12.7){S} \put(8.8,12.2){L 0.182 }
\put(8.8,10.7){S 0.818} \put(8.8,10){L} \put(8.8,8.7){S}


\put(7,18){\line(0,-1){1}} \put(7,15.5){\line(0,-1){1}}
\put(7,10.5){\line(0,-1){1}}

\put(6.5,18.8){$E$} \put(6.5,17.3){$\overline{E}$}
\put(6.5,15.){$E$} \put(6.5,13.5){$\overline{E}$}
\put(6.5,11.2){$E$} \put(6.5,9.7){$\overline{E}$}

\put(5,3.5){\line(0,-1){1}} \put(4,5){$E$}
\put(4,2.6){$\overline{E}$}

\put(7,6.8){L 0.1} \put(7,4.5){S 0.9} \put(7,3){L 0.1}
\put(7,.7){S 0.9}

\end{picture}

$P(L)=0.1,\ P(S)=0.9$ are prior probabilities.

From the table:

 $P(P|L)=0.6,\ P(N|L)=0.2\ P(I|L)=0.2,\\
P(P|S)=0.1,\ P(N|S)=0.8,\ P(I|S)=0.1$


\end{description}



\end{document}
