\begin{verbatim}
%http://www.iaw.com/~falls/origins.html#Power
%#http://blue.census.gov/cgi-bin/ipc/popclockw
http://www.wri.org/climate/carboncy.html
\end{verbatim}





\problem If two sports teams are almost equally skillful, then the
outcome of a single game between them might well not be the better
team; the weaker team might just get lucky.  To correct this problem,
the teams could play a best-of-five tournament.  This should allow the
stronger team to come back from an unlucky loss or two and triumph in
the end!

Let's test that hypothesis.  Suppose that Team A beats Team B in each
game with probability 0.51, regardless of the outcomes of other games.
We want to determine the probability that Team A beats Team B in a
best-of-five tournament.

\begin{problemparts}

\problempart What is the sample space for this experiment?

\problempart What subset of the sample space constitutes the event
that Team A wins the tournament?

\problempart What is the probability of each outcome in the sample space?

\problempart What is the probability that Team A wins the tournament?

\end{problemparts}

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\problem Suppose that you pick a pair of distinct numbers in the range
$1$ to $10$ uniformly at random.

\begin{problemparts}

\problempart If one of the numbers is even, then what is the
probability that the other number is even?

\solution{The sample space consists of the $\binom{10}{2} = 45$ pairs
of distinct numbers in the range $1$ to $10$.  Since the sample space
is uniform, each outcome has probability $\frac{1}{45}$.  With this
observation, we can compute the probability of an event by counting
the number of outcomes in the event and then multiplying by
$\frac{1}{45}$.

Let $E$ be the event that at least one number is even, and let $B$ be
the event that both numbers are even.  We must compute

\begin{eqnarray*}
\Pr(B \mid E)	& = &	\frac{\Pr(B \cap E)}{\Pr(E)} \\
		& = &	\frac{\Pr(B)}{\Pr(E)}
\end{eqnarray*}

The second equality holds because $B$ is contained in $E$; that is, if
both number are even, then at least one of the numbers is even.

Since there are 5 even numbers in the range 1 to 10, the event $B$
contains $\binom{5}{2} = 10$ outcomes and thus has probability $10
\cdot \frac{1}{45} = \frac{2}{9}$.  The event $E$ consists of all pairs of
numbers in the range 1 to 10 that are not both odd.  That is, event
$E$ contains $45 - \binom{5}{2} = 35$ outcomes, and thus has
probability $35 \cdot \frac{1}{45} = \frac{7}{9}$.

Substituting these probabilities into the above equation gives

\begin{eqnarray*}
\Pr(B \mid E)	& = &	\frac{\Pr(B)}{\Pr(E)} \\
		& = &	\frac{\frac{2}{9}}{\frac{7}{9}} \\
		& = &	\frac{2}{7}.
\end{eqnarray*}
}

\problempart If one of the numbers is a 2, then
what is the probability that the other number is even?

\solution{ Let $T$ be the event that one number is 2, and let $U$ be
the event that one number is 2 and the other is even.  We must compute

\begin{eqnarray*}
\Pr(U \mid T)	& = &	\frac{\Pr(U \cap T)}{\Pr(T)} \\
		& = &	\frac{\Pr(U)}{\Pr(T)}
\end{eqnarray*}

The event $T$ consists of 9 outcomes, since every pair must contain 2
and there are 9 choices for the other number ($1, 3, 4, \dots, 10$).
The event $U$ consists of 4 outcomes, since every pair must contain a
2 and there are 4 choices for the other number ($4, 6, 8, 10$).
Therefore, we have:

\begin{eqnarray*}
\Pr(U \mid T)	& = &	\frac{\Pr(U)}{\Pr(T)} \\
		& = &	\frac{\frac{4}{45}}{\frac{9}{45}} \\
		& = &	\frac{4}{9}.
\end{eqnarray*}
}

\end{problemparts}









\problem Two teams, the Alligators and the Bumpkins, play a
best-of-five tournament.  The Alligators win each game with
probability $\frac{3}{5}$, regardless of the outcomes of other games.
What is the probability that the Alligators win the tournament
overall, given that the Bumpkins win at least one game?

\solution{Let $T$ be the event that the Alligators win the
tournament, and let $G$ be the event that the Bumpkins win at least
one game.  We want to computes:

\begin{eqnarray*}
\Pr\{ T \mid G \} & = & \frac{\Pr\{ T \cap G \}}{\Pr\{ G\}}
\end{eqnarray*}

We use the usual four-step procedure: find the sample space, define
events of interest, compute outcome probabilities, compute event
probabilities.  These steps are carried out in the diagram below.

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{pset2-tree1.eps}}}
\bigskip

The probability of $T \cap G$, the event that the Alligtors win the
tournament and the Bumpkins win at least one game, is $0.144 + 0.144 +
0.144 = 0.648$.  The probability of event $G$ alone is $1 - 0.144 = $

 }

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