<<echo=FALSE, results=hide>>=
ok <- FALSE
while(!ok) {
  pe <- round(runif(1, 0.05, 0.15), digits = 2)
  per <- round(runif(1, 0.6, 0.8), digits = 2)
  pnenr <- round(runif(1, 0.6, 0.8), digits = 2)

  prob1 <- pe * per
  prob2 <- pe * (1 - per)
  prob3 <- (1 - pe) * (1 - pnenr)
  prob4 <- (1 - pe) * pnenr

  tab <- cbind(c(prob1, prob3), c(prob2, prob4))

  sol <- c(tab[1, 1] / sum(tab[, 1]), tab[1, 1] / sum(tab[1, ]),
    tab[2, 1] / sum(tab[2, ]), tab[1, 2] / sum(tab[1, ]))

  ok <- sum(tab) == 1 & all(tab > 0) & all(tab < 1)
}

tab2 <- cbind(rbind(tab, colSums(tab)), c(rowSums(tab), 1))
tab2 <- format(tab2 * 100, digits = 2, nsmall = 2)

sol <- round(100 * c(tab, colSums(tab), rowSums(tab), sum(tab)), digits = 2)
lab <- c("E \\cap R", "\\overline{E} \\cap R", "E \\cap \\overline{R}", "\\overline{E} \\cap \\overline{R}",
  "R", "\\overline{R}", "E", "\\overline{E}", "\\Omega")
@


\begin{question}
An industry-leading company seeks a qualified candidate for a management position.
A management consultancy carries out an assessment center which concludes in making
a positive or negative recommendation for each candidate: From previous assessments they know that
of those candidates that are actually eligible for the position (event $E$) $\Sexpr{per * 100}\%$
get a positive recommendation (event $R$). However, out of those candidates that are not eligible
$\Sexpr{pnenr * 100}\%$ get a negative recommendation. Overall, they know that only
$\Sexpr{pe * 100}\%$ of all job applicants are actually eligible.

What is the corresponding fourfold table of the joint probabilities? (Specify all entries in percent.)

\begin{tabular}{|c|cc|c|}
\hline
                 & ~$R$~           & ~$\overline{R}$~   & sum\\ \hline
~$E$~            & ~##ANSWER1##\%~ & ~##ANSWER3##\%~    & ~##ANSWER7##\%~ \\
~$\overline{E}$~ & ~##ANSWER2##\%~ & ~##ANSWER4##\%~    & ~##ANSWER8##\%~ \\ \hline
sum              & ~##ANSWER5##\%~ & ~##ANSWER6##\%~    & ~##ANSWER9##\%~ \\
\hline
\end{tabular}

<<echo=FALSE, results=tex>>=
answerlist(rep("", length(sol)))
@

\end{question}


\begin{solution}
Using the information from the text, we can directly calculate the following joint probabilities:
\begin{eqnarray*}
  P(E \cap R) & = &
    P(R | E) \cdot P(E) = \Sexpr{per} \cdot \Sexpr{pe} = \Sexpr{prob1} = \Sexpr{100 * prob1}\%\\
  P(\overline{E} \cap \overline{R}) & = &
    P(\overline{R} | \overline{E}) \cdot P(\overline{E}) = \Sexpr{pnenr} \cdot \Sexpr{1 - pe} = \Sexpr{prob4} = \Sexpr{100 * prob4}\%.
\end{eqnarray*}
The remaining probabilities can then be found by calculating sums and differences in the fourfold table:

\begin{tabular}{|c|cc|c|}
\hline
                 & ~$R$~                    & ~$\overline{R}$~         & sum \\ \hline
~$E$~            & ~{\bf \Sexpr{tab2[1, 1]}}~ & ~{\it \Sexpr{tab2[1, 2]}}~ & ~{\bf \Sexpr{tab2[1, 3]}}~ \\
~$\overline{E}$~ & ~{\it \Sexpr{tab2[2, 1]}}~ & ~{\bf \Sexpr{tab2[2, 2]}}~ & ~{\it \Sexpr{tab2[2, 3]}}~ \\ \hline
sum              & ~{\it \Sexpr{tab2[3, 1]}}~ & ~{\it \Sexpr{tab2[3, 2]}}~ & ~{\bf \Sexpr{tab2[3, 3]}}~ \\
\hline
\end{tabular}

<<echo=FALSE, results=tex>>=
answerlist(paste("$P(", lab, ") = ", format(sol), "\\%$", sep = ""))
@

\end{solution}


%% META-INFORMATION
%% \extype{cloze}
%% \exsolution{\Sexpr{paste(sol, collapse = "|")}}
%% \exclozetype{num|num|num|num|num|num|num|num|num}
%% \exname{fourfold}
%% \extol{0.05}
%% \exextra[numwidth,logical]{TRUE}
