\problemdata      
        {probability-intro-6} % latex-friendly label for the prob.
        {sample space, conditional probability} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 cp9W, S99 tut11-2} % Usage (list ones you are aware of).
        {Albert Meyer, S02} % Last revision info (author, date).

\begin{problem}
You are organizing a neighborhood census and instruct your census takers
to knock on doors and note the sex of any child that answers the knock.
Assume that there are two children in a household and that girls and boys
are equally likely to be children and to open the door.

A sample space for this experiment has outcomes that are triples whose
first element is either \texttt{B} or \texttt{G} for the sex of the elder
child, likewise for the second element and the sex of the younger child,
and whose third coordinate is \texttt{E} or \texttt{Y} indicating whether
the \emph{e}lder child or \emph{y}ounger child opened the door.  For
example, $(\mathtt{B},\mathtt{G},\mathtt{Y})$ is the outcome that the elder
child is a boy, the younger child is a girl, and the girl opened the door.

\bparts

\problempart Let \emph{T} be the event that the household has two girls,
and \emph{O} be the event that a girl opened the door.  List the outcomes
in \emph{T} and \emph{O}.

\solution{$T=\set{GGE,GGY}, O=\set{GGE,GGY,GBE,BGY}$}

\ppart What is the probability $\prcond{T}{O}$, that both children are
girls, given that a girl opened the door?
\solution{1/2}

\problempart Where is the mistake in the following argument?

\begin{quote}
If a girl opens the door, then we know that there is at least one girl in
the household.  The probability that there is at least one girl is
\begin{equation}
1 - \prob{\text{both children are boys}} = 1 - (1/2 \times 1/2) = 3/4.
\end{equation}
So,
\begin{eqnarray}
\lefteqn{\prcond{T}{\text{there is at least one girl in the household}}}\\
& = & \frac{\prob{T \intersect \text{there is at least one girl in the household}}}
{\pr{\text{there is at least one girl in the household}}}\\
& = & \frac{\prob{T}}{\pr{\text{there is at least one girl in the household}}}\\
& = & (1/4) / (3/4) = 1/3.
\end{eqnarray}
Therefore, given that a girl opened the door, the probability that there
are two girls in the household is \textup{1/3}.
\end{quote}

\solution{The argument is a correct proof that 
\[
\prcond{T}{\text{there is at least one girl in the household}} = 1/3.
\]
The problem is that the event, $H$, that the household has at least one girl,
namely,
\[
H \eqdef \set{\mathtt{GGE,GGY,GBE,GBY,BGE,BGY}},
\]
is not equal to the event, \emph{O}, that a girl opens the door.  These
two events differ:
\[
H-O = \set{\mathtt{BGE,GBY}},
\]
and their probabilities are different.  So the fallacy is in the final
conclusion where the value of $\prcond{T}{H}$ is taken to be the same as
the value $\prcond{T}{O}$.  Actually, $\prcond{T}{O} = 1/2$.  }

\eparts
\end{problem}
