\documentclass[12pt,twoside]{article}
\usepackage{psfig}

\input{macros-course}

\begin{document}

\begin{problems}

\begin{problem}
There is a course---not 6.042, naturally---in which $10 \%$ of the
assigned problems contain errors.  If you ask a TA whether a problem has
an error, then he or she will answer correctly $80\%$ of the time.  This
$80 \%$ accuracy holds regardless of whether or not a problem has an
error.  Likewise when you ask a lecturer, but with only $75 \%$ accuracy.

\bparts

\ppart
Suppose you have doubts about a problem and ask a TA about it, and she
tells you that the problem is correct.  To double-check, you ask a
lecturer, who says that the problem has an error.  Assuming that \emph{the
correctness of the lecturers' answer and the TA's answer are independent
of each other, regardless of whether there is an error}\footnote{This
assumption is questionable: by and large, we would expect the lecturer and
the TA's to spot the same glaring errors and to be fooled by the same
subtle ones.}, what is the probability that there is an error in the
problem?

We formulate this as an experiment of choosing one problem randomly and
asking a particular TA and Lecturer about it.  Define the following events:
\begin{eqnarray*}
E & ::= & \mbox{``the problem has an error,''}\\
T & ::= & \mbox{``the TA says the problem has an error,''}\\
L & ::= & \mbox{``the lecturer says the problem has an error.''}
\end{eqnarray*}

\solution{The assumptions above tell us:
\begin{eqnarray*}
\pr{E}        & = & \frac{10}{100} = \frac{1}{10},\\
\prcond{T}{E} & = & \prcond{\bar{T}}{\bar{E}} = \frac{80}{100} = \frac{4}{5},\\
\prcond{L}{E} & = & \prcond{\bar{L}}{\bar{E}} = \frac{75}{100} = \frac{3}{4},
\end{eqnarray*}

Also, $T$ and $L$ are independent given $E$, and given $\bar{E}$:
\begin{eqnarray*}
\prcond{T \intersect L}{E} & = & \prcond{T}{E} \prcond{L}{E}\\
\prcond{T \intersect L}{\bar{E}} & = & \prcond{T}{\bar{E}} \prcond{L}{\bar{E}}
\end{eqnarray*}
Note that while we know that $T$ and $L$ are independent \emph{given} $E$
or \emph{given} $\bar{E}$, they are not independent by themselves, \cf
part~(\ref{TL}).

We want to calculate
\[
\prcond{E}{\bar{T} \intersect L}.
\]

>From the definition of conditional probability (this is Bayes' rule):
\begin{equation}\label{eq:B2}
 \prcond{E}{\bar{T} \intersect L} = \pr{E}
 \frac{\prcond{\bar{T} \intersect L}{E}}
      {\pr{\bar{T} \intersect L}}.
\end{equation}

By the independence assumptions, we have:
\begin{eqnarray*}
\prcond{\bar{T} \intersect L}{E} & = & \prcond{\bar{T}}{E} \prcond{L}{E}
     = \frac{1}{5}\frac{3}{4}  = \frac{3}{20},\\
\prcond{\bar{T} \intersect L}{\bar{E}} & = &
         \prcond{\bar{T}}{\bar{E}} \prcond{L}{\bar{E}}
     = \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5},\\
\pr{\bar{T} \intersect L} & = & \prcond{\bar{T} \intersect L}{E}  \pr{E} + 
                          \prcond{\bar{T} \intersect L}{\bar{E}} \pr{\bar{E}}\\
 & = & \frac{3}{20} \cdot \frac{1}{10} + \frac{1}{5} \cdot \frac{9}{10}
       = \frac{39}{200}.
\end{eqnarray*}

Substituting these values in equation~(\ref{eq:B2}), we get
\[
\prcond{E}{\bar{T} \intersect L} = \frac{1}{10} \cdot \frac{3/20}{39/200} =
\frac{1}{13} \approx 0.077.
\]
So this contradictory information has decreased the probability of an error
from $10 \%$ to about $7.7 \%$.

The calculations here support the common-sense rule that when two people
make contradictory statements, you should be influenced more by the most
``authoritative'' person---the one who is right more often.  But note that
this does not mean that you should {\em believe} in what the most
authoritative person says, since the probability of an error remains
uncomfortably large.}

\ppart\label{TL} Is the event that ``the TA says that there is an error'',
independent of the event that ``the lecturer says that there is an
error''?

\solution{The answer is no.  Because the TA is usually right, when the TA
says that the problem has an error, the likelihood that there really is an
error is increased.  But the lecturer is also usually right, so increasing
the likelihood of there \emph{being} an error also increases the
likelihood that the lecturer will \emph{report} an error.

We verify this informal argument by actually calculating the probability
of each of these events and their conjunction, and observing that the
probability that the two events occur is different from the product of the
probabilities.  Let events $E,T,L$ be as above.

\begin{eqnarray*}
\pr{T} & = & \pr{T \cap E} + \pr{T \cap \bar{E}}\\
 & = & \pr{T \mid E} \pr{E}  + \pr{T \mid \bar{E}} \pr{\bar{E}}\\
 & = & \frac{4}{5} \frac{1}{10} + (1 - \frac{4}{5})
 (1 - \frac{1}{10}) = \frac{13}{50},\\
& & \\
\pr{L} & = & \pr{L \cap E} + \pr{L \cap \bar{E}}\\
  & = & \frac{3}{4} \frac{1}{10} + (1 -\frac{3}{4})(1 - \frac{1}{10}) = \frac
{3}{10},\\
& & \\
\pr{L \cap T} & = & \pr{L \cap T \cap E} + \pr{L \cap T \cap \bar{E}} \\
& = & \pr{L \cap T \mid  E} \pr{E} + \pr{L \cap T \mid  \bar E} \pr{\bar E}\\
& = & \pr{L \mid E}      \pr{T \mid  E}       \pr{E} + 
      \pr{L \mid \bar E} \pr{T \mid  \bar{E}} \pr{\bar E}\\
& = & \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{1}{10}
        + (1 - \frac{3}{4}) (1 - \frac{4}{5}) \cdot (1 - \frac{1}{10}) = \frac
{105}{1000} = 0.105,
\end{eqnarray*}
which is higher than 
\[
\pr{L}\pr{T} = \frac{3}{10} \cdot \frac{13}{50} = .078.
\]
}

\eparts
\end{problem}

\end{problems}

\end{document}


