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\begin{document}
\pset{10}{November 21, 2000}{2:35pm in lecture, Tuesday, November 28, 2000}

\textbf{Be sure to indicate your TA's name clearly on the solutions
you turn in}.

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\textbf{Self-study}
Read: Lecture notes, Rosen Sections 4.4 and 4.5.\\ 
Self-study problems (not to be handed in): 4.4: 13, 33. 4.5: 19, 23.\\
%[[[Miniquiz for tut5: 11]]]
--------------------------------------------------------------------------------------------------------------\\
\paragraph{Policy on collaboration}
If you collaborated with someone on the problem sets, you should mention their names on top of the problem set. If you did not collaborate with anyone, you should explicitly say so. As a reminder, the write-up of your solutions should be entirely your own.
 
\medskip
\begin{problems}

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


\problem 
% 18.313 Quiz I, problem 4. 

Suppose that 5 men out of 100, and 25 women out of 10,000 are
colorblind.  A colorblind person is chosen at random from a population
with equal numbers of males and females.  What is the probability of
this person being male?

\solution{
\medskip
\textbf{Solution:}
\medskip
Let M be the event the person is male, F female,
C colorblind.  We are given $\prob{C \mid M} = \frac{5}{100}$ and
$\prob{C \mid F}=\frac{25}{10,000}.$ Since the population consists of equal
numbers of males and females, $\prob{M} = \prob{F} = \frac{1}{2}$.  So,
\[\prob{M \mid C} = \frac{\prob{C \mid M}\times \prob{M}}{\prob{C}} = \frac{\prob{C \mid M}\times \prob{M}}{\prob{C \mid M}\times \prob{M} + \prob{C \mid F}\times \prob{F}} = \]
\[\frac{(5/100)\times (1/2)}{(5/100)\times (1/2) + (25/10000) \times
(1/2)} = 20/21. \]}

\problem
Suppose that a resident of Boston chosen at random is a Red Sox fan (The Curse
 notwithstanding) with probability $0.90$. Also suppose that a 
resident chosen at
random is a Celtics fan (probably still living in the Bird-McHale-Parish era)
with probability $0.75$, and that a resident
Red Sox fan chosen at random is a Celtics fan with probability
$0.80$.  Now suppose that a resident is chosen at random from among
those that are {\it not\/} Red Sox fans.  What is the probability
that such a resident is a Celtics fan?

\solution{
\medskip
\textbf{Solution:}
\medskip
Let $C$ represent the event that a resident is a Celtics fan, and $S$ represent the event
that the resident is a Red Sox fan.  Then:
 
\begin{eqnarray*}
\prob{C} & = & \prob{C  \mid  S} \prob{S} + \prob{C  \mid  \overline{S}} \prob{\overline{S}} \\
\prob{C  \mid  \overline{S}} & = & \frac{\prob{C} - \prob{C  \mid  S} * \prob{S}}{\prob{\overline{S}}} \\
& = & \frac{0.75 - 0.80 \cdot 0.90}{1 - 0.9} \\
& = & 0.3
\end{eqnarray*}}
\problem 
% 18.313 Quiz I, problem 3

A population consists of $c_1$ members of type 1, $c_2$ members
of type $2,\cdots ,c_n$ members of type $n$.  A set of $k$ members is
drawn from the population without replacement. 

\bparts
\ppart Find the probability that exactly $i$ members of type 1
were chosen.

\solution{
\medskip
\textbf{Solution:}
\medskip
Let $N_{1i}$ be the event that exactly $i$ members of class 1 were
chosen.  $\prob{N_{1i}} = \frac{\binom{c_1}{i} \times \binom{c_2+\cdots +
c_n}{k-i}}{\binom{c_1+\cdots +c_n}{k}}$.

}

\ppart Find the probability that exactly $i$ members of class 1
and exactly $j$ members of class 2 were chosen.

\solution{
\medskip
\textbf{Solution:}
\medskip
Let $N_{2j}$ be the event that exactly $j$ members of class 2 were
chosen.  Then $ \prob{N_{1i} \cap N_{2j}} = \frac{\binom{c_1}{i} \binom{c_2}{j}
\binom{c_3 +\cdots + c_n}{k-i-j}}{\binom{c_1 + \cdots +c_n}{k}} $}

\eparts


\problem
In London, half of the days have some rain. The weather forecaster is
correct 2/3 of the time -- i.e. the probability that it rains, given
that she has predicted that it will rain, and the probability that it
does not rain given that she has predicted that it does not rain, are
both equal to 2/3. When rain is forecast, Mr. Pickwick takes his
umbrella. When rain is not forecast, he takes it with probability
1/3. 
\bparts
\ppart Find the probability that Pickwick has no umbrella, given that it rains.
\ppart Find the probability that it doesn't rain, given that he brings his umbrella.  
\eparts

\solution{
\medskip
\textbf{Solution:}
\medskip 
\\We construct a tree as in lecture notes \#19. For part (a), we wish to find 
\begin{eqnarray*}
\prob{\text{no umbrella}  \mid \text{rain}}
  & = &\frac{\prob{\mbox{no umbrella} \wedge \mbox{rain}}}{\prob{\mbox{rain}}}\\
& = & \frac{1/9}{1/2} = 2/9.
\end{eqnarray*}

For part (b), we wish to find
\begin{eqnarray*}
\prob{\text{no rain} \mid \text{umbrella}} & = &
 \frac{\prob{\mbox{no rain} \wedge \mbox{umbrella}}}{\prob{\mbox{umbrella}}}\\
& = & \frac{1/6 + 1/9}{1/9 + 1/6 + 1/3 + 1/18} = 5/12
\end{eqnarray*}

}

\problem

Let $A,B,C$ be events of some experiment.  For each of the following
statements, prove it if it is true or give a counterexample if it is
false:

\bparts

\ppart If $C$ is independent of $A$, and $C$ is independent of $B$, then
$C$ is independent of $A\cup B$.

\solution{
\medskip
\textbf{Solution:}
\medskip
The statement is false.  For example, let $A$ and $B$ be
unbiased independent bits, and let $C = A \oplus B$.  Then $C$ is
independent from both $A$ and $B$, but it is not independent from $A
\cup B = A\, {\sf or}\, B$}.

\ppart If $C$ is independent of $A$, and $C$ is independent of $B$, and
$C$ is independent of $A\cap{B}$, then $C$ is independent of $A\cup{B}$.

\solution{
\medskip
\textbf{Solution:}
\medskip
It is true.  To prove it, observe by the Sum Rule that
\begin{eqnarray*}
\lefteqn{\prob{x\in(C\cap(A\cup B))}}\\
& = &  \prob{x\in(C\cap(A- B))} +\prob{x\in(C\cap(B- A))} + \prob{x\in(C\cap(A\cap B))},
\end{eqnarray*}
and then that 
\[
\prob{x\in(C\cap(A - B))} + \prob{x\in(C\cap (A \cap B))}
= \prob{x \in (C\cap A)}
\]
also by the Sum Rule.  Then use the fact that $C$ is independent of
both $A$ and $A \cap B$ to get that $C$ is independent from $A - B$, and
similarly that it is independent from $B - A$.

}
\eparts


\problem Compute the probabilities of the following
three events and determine which one is most likely.  Assume that the
dice are fair, 6-sided, and mutually independent.

\begin{problemparts}
\problempart Rolling at least one 6 when six dice are rolled.

\solution{
\medskip
\textbf{Solution:}
\medskip
The probability of rolling a 6 on a particular die is
$\frac{1}{6}$.  The probability of not rolling a 6 on a particular die
is $\frac{5}{6}$, since this is the complementary event.  The
probability of not rolling a 6 on any of six dice is
$(\frac{5}{6})^6$, since the numbers rolled are mutually independent.
The probability of rolling a 6 on some die is therefore $1 -
(\frac{5}{6})^6 = 0.665\ldots$.  }

\problempart Rolling at least two 6's when twelve dice are rolled.

\solution{
\medskip
\textbf{Solution:}
\medskip
The probability of rolling two or more 6's is 1 minus the
probability of rolling zero or one 6's.  The probability of rolling
zero 6's is $(\frac{5}{6})^{12}$, arguing as before.  The probablity
of rolling exactly one 6 is

\[
12 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{11}
\]

since there are 12 ways to choose the die that comes up 6, the
probability of this die coming up 6 is $\frac{1}{6}$, and the
probability of the remaining dice all coming up in the range 1 to 5 is
$(\frac{5}{6})^{11}$.

Overall, the probability of rolling two or more sixes is therefore

\begin{eqnarray*}
1 - \left(\frac{5}{6}\right)^{12} -
	12 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{11}
		& = &	0.618\ldots.
\end{eqnarray*}
}

\problempart Rolling at least three 6's when eighteen dice are rolled.

\solution{
\medskip
\textbf{Solution:}
\medskip
Arguing as before, the probability of rolling at least three 6's is

\begin{eqnarray*}
1	- \left(\frac{5}{6}\right)^{18}
	- 18 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{17}
	- \binom{18}{2} \cdot \left(\frac{1}{6}\right)^2 \cdot
		\left(\frac{5}{6}\right)^{16}
	& = &	0.597\ldots.
\end{eqnarray*}
}

\end{problemparts}

\solution{
\medskip
\textbf{Solution:}
\medskip
The first event is most likely.}

\end{problems}
\end{document}
