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  \paragraph{Problem}}
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  \paragraph{Solution}}

\begin{document}


\begin{problems}

%BEGIN Kavita's problems
% Fall 1998


%**************
%   PSET 9    *
%**************


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\problem {\bf (10 points)} In each problem part, either define a probability
space and events $A$, $B$, and $C$ such that all the equations hold
simultaneously, or else prove that no such probability space and
events exist.

\begin{problemparts}

\problempart
\begin{eqnarray*}
\Pr(A \mid B)		& = &	\frac{1}{4} \\
\Pr(A \mid C)		& = &	\frac{1}{4} \\
\Pr(A \mid B \cap C)	& = &	\frac{3}{4}
\end{eqnarray*}

\solution{These events can exist, as indicated in the diagram below.
Empty regions have zero probability.

\bigskip
\centerline{\resizebox{!}{1.5in}{\includegraphics{Sol9-set1.eps}}}
\bigskip
}

\problempart
\begin{eqnarray*}
\Pr(A \mid B)		& = &	\frac{3}{4} \\
\Pr(A \mid C)		& = &	\frac{3}{4} \\
\Pr(A \mid B \cap C)	& = &	\frac{1}{4}
\end{eqnarray*}

\solution{These events can exist, as indicated in the diagram below.

\bigskip
\centerline{\resizebox{!}{1.5in}{\includegraphics{Sol9-set2.eps}}}
\bigskip
}

\problempart
\begin{eqnarray*}
\Pr(A \mid B)		& = &	\frac{3}{4} \\
\Pr(A \mid C)		& = &	\frac{3}{4} \\
\Pr(A \mid B \cup C)	& = &	\frac{1}{4}
\end{eqnarray*}

\solution{These events can not exist.  At least one of the events $B$ or
$C$ must be at least half as likely as $B \cup C$.  Without loss of
generality, assume that this is true for event $B$; that is,

\begin{eqnarray*}
\frac{\Pr(B)}{\Pr(B \cup C)}	& \geq &	\frac{1}{2}.
\end{eqnarray*}

The first equation implies that

\begin{eqnarray*}
\frac{\Pr(A \cap B)}{\Pr(B)}	& = &		\frac{3}{4}.
\end{eqnarray*}

Multiplying the equation and the inequality above gives

\begin{eqnarray*}
\frac{\Pr(A \cap B)}{\Pr(B \cup C)}	& \geq &	\frac{3}{8}.
\end{eqnarray*}

Note that the set $A \cap (B \cup C)$ contains the set $A \cap B$.
Therefore, $\Pr(A \cap (B \cup C)) \geq \Pr(A \cap B)$.  Substituting
this fact into the preceding inequality gives

\begin{eqnarray*}
\frac{\Pr(A \cap (B \cup C))}{\Pr(B \cup C)}	& \geq &	\frac{3}{8}.
\end{eqnarray*}

Consequently, $\Pr(A \mid B \cup C) \geq \frac{3}{8} > \frac{1}{4}$,
contradicting the third equation.
}

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem {\bf (10 points)}  After their arrest and sentencing for
unlawful use of firearms, and short hospital stays, the three
gunfighters, $A$, $B$, and $C$ applied for parole.  The parole board
decided to release two of the three, chosen uniformly at random, but
has not released their names.

Gunfighter $A$ figures that he will be released with probability
$\frac{2}{3}$.  A guard offers to tell gunfighter $A$ the name of one
of the other gunfighters who will be released (either $B$ or $C$).
However, gunfighter $A$ declines the offer.  He reasons that if the
guard says, for example, ``$B$ will be released'', then his own
probability of release will drop to $\frac{1}{2}$.  This is because he
will then know that either he or $C$ will also be released, but these
events are equally likely.

Either prove that gunfighter $A$ has reasoned correctly or prove that
he is wrong.  Assume that if the guard has a choice of saying $B$ or
$C$ because both are to be released, then he picks one uniformly at
random.

\solution{Gunfighter $A$ has reasoned incorrectly.  We must compute

\begin{eqnarray*}
\Pr(\text{$A$ released} \mid \text{guard says ``$B$''})
	& = &	\frac{\Pr(\text{$A$ released and guard says ``$B$''})}
			{\Pr(\text{guard says ``$B$''})}.
\end{eqnarray*}

These probabilities are worked out in the tree diagram below.

\bigskip
\centerline{\resizebox{!}{3.0in}{\includegraphics{Sol9-prisoner.eps}}}
\bigskip

Plugging these probabilities into the preceding equation gives

\begin{eqnarray*}
\Pr(\text{$A$ released} \mid \text{guard says $B$ released})
	& = &	\frac{\frac{1}{3}}{\frac{1}{2}} \\
	& = &	\frac{2}{3}
\end{eqnarray*}

If the guard says that $B$ will be released, then $A$ still has a
$\frac{2}{3}$ chance of release.
}

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%END   Kavita's problems

%BEGIN Joe's problems
%END   Joe's problems

%BEGIN Ray's problems
%END   Ray's problems

%BEGIN Ivan's problems (spring '98)
%END   Ivan's problems

%BEGIN George's problems
%END   George's problems

%BEGIN Josh's problems
%END   Josh's problems

%BEGIN Tina's problems
%END   Tina's problems


\end{problems}

\end{document}



