\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Conditional} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, quiz 3} % Usage (list ones you are aware of).
        {Christos Kapoutsis, 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.
\begin{problemparts}

\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
\begin{itemize}
\item \emph{T}: \brule{5in}  \solution{GGE,GGY}
\item \emph{O}: \brule{5in} \solution{GGE,GGY,GBE,BGY}
\end{itemize}

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

\problempart 
Where is the mistake in the following argument?

\begin{quotation}
{\em
\noindent 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{quotation}

\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 $\prcond{T}{\text{there is at least one girl
in the household}}$ is taken to be the same as the value $\prcond{T}{O}$.
Actually, $\prcond{T}{O} = 1/2$.

\iffalse
For having children, each case GG, GB, BG and BB is equally
likely.  However given GG, the probability that a girl opens the door
is 1.  Given the GB or BG the probability is 1/2.  Given BB the
probability that a girl opens the door is 0.  Therefore the probability
that a girl opens the door is not the same for all four cases of
children pairs.  Since the four children pairs are disjoint we can
simply add up the conditional probabilities that a girl opens for each
pair (by the total probability rule).

Note that GB and BG are distinct cases. It is similar to saying that
flipping HT is different from flipping TH for after all what is having
kids but flipping a coin.
\fi
}

\end{problemparts}
\end{problem}
