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\usepackage{latex-macros/course}

%\hidesolutions
\showsolutions

\begin{document}

\renewcommand{\reading}{Course Notes 10.}

\problemset{9}

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\begin{problem}
Suppose that you flip four fair, independent coins.  The goal of this
problem is to determine which of the following two events is more
likely:

\begin{enumerate}

\item Two coins come up one way and two come up the other (as in $HTTH$).

\item Three coins come up one way and one comes up the other (as in
$TTHT$ and $HHHT$).

\end{enumerate}

We will use the four-step method described in the Course Notes.  Draw
a tree diagram as you work through the questions below.  {\em Include
the tree diagram with your solution}.

\solution{
A tree diagram for this experiment is shown below.

\bigskip
\centerline{\includegraphics[height=4in]{rec7-cointree}}
\bigskip
}

\bparts

\ppart Step 1: What is the sample space for this experiment?

\solution{
Since this experiment has such a regular structure, we'll
dispense with the tree diagram.

The sample space consists of sixteen outcomes, each of which specifies
the outcomes of the four coin flips:

\[
\begin{array}{lccccr}
\{ & HHHH, & HHHT, & HHTH, & HHTT, \\
   & HTHH, & HTHT, & HTTH, & HTTT, \\
   & THHH, & THHT, & THTH, & THTT, \\
   & TTHH, & TTHT, & TTTH, & TTTT & \}
\end{array}
\]

The outcome $HHHT$, for example, corresponds to the case where the
first three coins come up heads and the last comes up tails.
}

\ppart Step 2: What subsets of this sample space constitute the two
events listed above?

\solution{
The first event consists of outcomes containing an equal number of
$T$'s and $H$'s.  There are six of these:

\[
\{ HHTT, HTHT, HTTH, THHT, THTH, TTHH \}
\]

The second event consists of outcomes containing three of one
orientation and one of the other.  There are eight of these:

\[
\{ HTTT, THTT, TTHT, TTTH, THHH, HTHH, HHTH, HHHT \}
\]
}

\ppart Step 3: What is the probability of each outcome in the sample
space?

\solution{
Each outcome has probability $1/16$.
}

\ppart Step 4: What are the probabilities of the two events?

\solution{
Recall that the probability of an event is defined to be
the sum of the probabilities of the outcomes in that event.
Therefore, we have:

\begin{eqnarray*}
\lefteqn{\Pr(\text{two one way, two the other})} \hspace{0.5in} \\
	& = &	\Pr\{HHTT\} + \Pr\{HTHT\} + \Pr\{HTTH\} \\
        &   &	+ \Pr\{THHT\} + \Pr\{THTH\} + \Pr\{TTHH\} \\
	& = &	\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + 
		\frac{1}{16} + \frac{1}{16} + \frac{1}{16} \\
	& = &	\frac{3}{8}
\end{eqnarray*}

\begin{eqnarray*}
\lefteqn{\Pr(\text{three one way, one the other})} \hspace{0.5in} \\
	& = &	\Pr\{HTTT\} + \Pr\{THTT\} + \Pr\{TTHT\} + \\
	& &	\Pr\{TTTH\} + \Pr\{THHH\} + \Pr\{HTHH\} + \\
	& &	\Pr\{HHTH\} + \Pr\{HHHT\} \\
	& = &	\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + 
		\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} \\
	& = &	\frac{1}{2}
\end{eqnarray*}
}

\ppart Which event is more likely?

\solution{
The second event, that three coins come up one way and one
comes up the other, is more likely.
}

\eparts

\end{problem}

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\begin{problem}
Outside of their hum-drum duties at 6.042 TAs, Min is trying to learn
to levitate using only intense concentration and Mana is trying to
become the world heavyweight boxing champion.  Suppose that Min's
probability of success is $1/6$, Mana's chance of success is $1/4$,
and these two events are independent.

\bparts

\ppart If at least one of them succeeds, what is the probability that
Min learns to levitate?

\solution{Let $L$ be the event that Min learns to levitate, and let
$B$ be the event that Mana becomes the boxing champion.  We can work
out the desired probability as follows:

\begin{eqnarray*}
\pr{L \mid (L \cup B)}
	& = &	\frac{\pr{L \cap (L \cup B)}}{\pr{L \cup B}} \\
	& = &	\frac{\pr{L}}{1 - \pr{\overline{L} \cap \overline{B}}} \\
	& = &	\frac{1/6}{1 - (1 - 1/6)(1 - 1/4)} \\
	& = &	\frac{4}{9}
\end{eqnarray*}

The first step uses the definition of conditional probability.  In the
second step, we rewrite both the top and bottom of the fraction using
set identities.  Then we substitute in the given probability and
simplify.}

\ppart If at most one of them succeeds, what is the probability that
Mana becomes the world boxing champion?

\solution{
Define events $L$ and $B$ as before.

\begin{eqnarray*}
\pr{B \mid (\overline{L} \cup \overline{B})}
	& = &	\frac{\pr{B \cap (\overline{L} \cup \overline{B})}}
			{\pr{\overline{L} \cup \overline{B}}} \\
	& = &	\frac{\pr{B \cap \overline{L}}}
			{1 - \pr{L \cap B}} \\
	& = &	\frac{(1/4) \cdot (5/6)}{1 - (1/6) \cdot (1/4)} \\
	& = &	\frac{5}{23}
\end{eqnarray*}
}

\ppart If exactly one of them succeeds, what is the probability that
it is Min?

\solution{
\begin{eqnarray*}
\pr{L \mid ((L \cap \overline{B}) \cup (\overline{L} \cap B))}
	& = &	\frac{\pr{L \cap \overline{B}}}
			{\pr{((L \cap \overline{B}) \cup
			(\overline{L} \cap B))}} \\
	& = &	\frac{(1/6) \cdot (3/4)}
			{(1/6) \cdot (3/4) + (5/6) \cdot (1/4)} \\
	& = &	\frac{3}{8}
\end{eqnarray*}
}

\eparts
\end{problem}

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\begin{problem}
Suppose that you roll a fair die that has $n$ sides, numbered $1, 2,
\ldots, n$.  Determine whether the event that you roll a multiple of 2
is independent of the event that you roll a multiple of 3 when:

\bparts

\ppart $n = 6$

\solution{Let $A$ be the event that the number on top is a multiple of
2, and let $B$ be the event that the number on top is a multiple of 3.
We have:

\[
\pr{A} \cdot \pr{B} =
	\frac{3}{6} \cdot \frac{2}{6} =
	\frac{1}{6} = \pr{A \cap B}
\]

Therefore, these events are independent.
}

\ppart $n = 4$

\solution{As before, let $A$ be the event that the number on top is a
multiple of 2, and let $B$ be the event that the number on top is a
multiple of 3.  Now, however, we have:

\begin{eqnarray*}
\pr{A} \cdot \pr{B}
	& = &	\frac{2}{4} \cdot \frac{1}{4} \\
	& = &	\frac{1}{8}
\end{eqnarray*}

But:

\begin{eqnarray*}
\pr{A \cap B} & = & 0
\end{eqnarray*}

Since these results disagree, the events are not independent.
}

\ppart $n = 8$

\solution{
Now we have:

\[
\pr{A} \cdot \pr{B} =
	\frac{4}{8} \cdot \frac{2}{8} =
	\frac{1}{8} = \pr{A \cap B}
\]
}

\eparts

Justify each of your answers.

\end{problem}

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\begin{problem}
There are three prisoners in a maximum-security prison for fictional
villains: the Evil Wizard Voldemort, the Dark Lord Sauron, and Little
Bunny Foo-Foo.  The parole board has declared that it will release two
of the three, chosen uniformly at random, but has not yet released
their names.  Naturally, Sauron figures that he will be released to
his home in Mordor, where the shadows lie, with probability
$\frac{2}{3}$.

A guard offers to tell Sauron the name of one of the other prisoners
who will be released (either Voldemort or Foo-Foo).  However, Sauron
declines this offer.  He reasons that if the guard says, for example,
``Little Bunny Foo-Foo 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 Voldemort will also be released, and these two
events are equally likely.

Using conditional probability, either prove that the Dark Lord Sauron
has reasoned correctly or prove that he is wrong.  Assume that if the
guard has a choice of naming either Voldemort or Foo-Foo (because both
are to be released), then he names one of the two uniformly at random.
\end{problem}

\solution{Sauron has reasoned incorrectly.  His mistake is not
realizing that the event ``Foo-Foo will be released'' is different
from the event ``The guard says Foo-Foo will be released''.  The tree
diagram below shows that those two events are different sets of
outcomes.  The first split in the tree is based on which two of the
three prisoners are to be released, and the second is based on what
the guard tells Sauron.

\bigskip
\centerline{\resizebox{!}{2.8in}{\includegraphics{prison.eps}}}
\bigskip

Taking probabilities from the tree diagram, we can compute Sauron's
true probabiliy of release as follows:

\begin{eqnarray*}
\lefteqn{
\pr{\text{Sauron released} \mid \text{guard says ``Foo-Foo''}}} \hspace{1in} \\
	& = &	\frac{\pr{\text{Sauron released} \cap
			\text{guard says ``Foo-Foo''}}}
		{\pr{\text{guard says ``Foo-Foo''}}} \\
	& = &	\frac{1/3}{1/3 + 1/6} \\
	& = &	\frac{2}{3}
\end{eqnarray*}

Therefore, if the guard says that Little Bunny Foo-Foo will be
released, then Sauron, Lord of Mordor, still has a $\frac{2}{3}$
chance of release.
}

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\begin{problem}
A 52-card deck is thoroughly shuffled and you are dealt a hand of 13
cards.

\bparts

\ppart If you have one ace, what is the probability that you have a
second ace?

\solution{
}

\ppart If you have the ace of spades, what is the probability that you
have a second ace?

\solution{
}

\eparts

Remarkably, the two answers are different.  This problem will test
your counting ability!

\end{problem}

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\begin{problem}
Let's play a game!  We repeatedly flip a fair coin.  You have the
sequence $HHT$, and I have the sequence $HTT$.  If your sequence comes
up first, then you win.  If my sequence comes up first, then I win.
For example, if the sequence of tosses is:

\[
TTHTHT\underline{HHT}
\]

\noindent then you win.  This problem is tricky, because the game
could go on for an arbitrarily long time.  Draw enough of the tree
diagram to see a pattern, and then sum up the probabilities of the
(infinitely many) outcomes in which you win.

\solution{A partial tree diagram is shown below.  All edge
probabilities are $1/2$.

\bigskip
\centerline{\includegraphics[height=3in]{ps9-HHTtree}}
\bigskip

Let's first focus on the subtree shown in bold.  Note that if two
heads are flipped in a row, then you are guaranteed to win eventually.
The sum of the probabilities of all your winning outcomes in this
subtree is:

\begin{eqnarray*}
\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots
	& = & \frac{1}{4} \cdot \frac{1}{1 - 1/4} \\
	& = & \frac{1}{3}
\end{eqnarray*}

The upppermost subtree marked {\bf same} is the identical to the one
shown in bold, except that each outcome probability is reduced by
$1/2$, because it is one edge farther from the root.  Thus, the sum of
your winning outcomes in this subtree is $1/6$.  Similarly, the sum of
your winning outcomes in the next subtree marked {\bf same} is $1/12$,
and so forth.  Overall, your probability of winning is:

\begin{eqnarray*}
\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \ldots
	& = & \frac{1}{3} \cdot \frac{1}{1 - 1/2} \\
	& = & \frac{2}{3}
\end{eqnarray*}
}
\end{problem}

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\end{document}
