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\begin{document}

\psol{11}{November 26, 1996}

\bigskip

\begin{problems}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem

Total number of hands is ${52 \choose 13}$.
To count the number of hands with 9 cards of the same suit,
we need to (1) choose the suit, (2) choose the 9 cards of the same suit,
and (3) choose the remaining 4 cards.
This can be done in $4{13 \choose 9}{39 \choose 4}$ ways.
Note, there is no possible double counting, since
a hand can have at most one 9-carded suit.
Thus, the probability is 
\[
\frac{4{13 \choose 9}{39 \choose 4}}{{52 \choose 13}} = 0.0003704 
%235237860/635013559600
\]

The number of hands with exactly 5 spades is ${13 \choose 5}{39 \choose 8}$,
which is the same as the number of hands with exactly 5 hearts.
However, we double counted.   The number of hands with both 5 spades
and 5 hearts is ${13 \choose 5}{13 \choose 5}{26 \choose 3}$.
Thus, the probability is 
\[
\frac{2{13 \choose 5}{39 \choose 8}-{13 \choose 5}{13 \choose 5}{26 \choose 3}}
{{52 \choose 13}} = 0.2426 %158362127352- 4306559400/635013559600=
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\begin{problemparts}
\problempart

For a given ticket, it will win if any of ${20 \choose 10}$ sets of
numbers are choosen.  Since there are ${80 \choose 10}$ equally likely
sets of numbers, the probability of winning is
\[
\frac{{20 \choose 10}}{{80 \choose 10}} = 0.0000001122
%184756/1646492110120
\]
We will denote this probability by ${\cal P}$.

\problempart
For the lower bound, if all of the students select the same
set of numbers,  then the probability of winning is the same as above.
Thus, the lower bound is ${\cal P}$.

For the upper bound, 
the probability of any student winning is at most the sum of
probabilities of each student winning, which is $142 {\cal P}$.


\ppart
Here, we are only concerned about the number of distinct tickets.
For an upper bound, the probability of a win is at
most the sum of the probabilities of each distinct ticket of winning
-- $k{\cal P}$.

The lower bound is more difficult to analyze.  We need to analyze
the number of distinct sets of ten numbers.  A simple lower bound
is probability of a single ticket winning -- ${\cal P}$.

\ppart
All the students could have choosen the same ticket.  In that case, if one
student wins, they all win.  Thus, a lower bound on the probability
of exactly one winner is zero.  On the other hand, the tickets
could be arranged so that no two different tickets have the same 10 numbers
choosen.  In that case, if exaclty one person can win at any given time,
the probability of a win is as in (b), $142{\cal P}$.

\ppart
Let $s$ be the number of distinct tickets chosen by only
one person (i.e. $p_i=1$).  For there to be exactly one winner,
one of these $s$ tickets must be chosen and no other ticket wins.
This probability has an upper bound of $s{\cal P}$.
For the lower bound, it is possible that every set of 10 numbers
appears on two different tickets.  Thus, the lower bound is zero.

\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\begin{problemparts}
\problempart
Out of the $n$ tosses, $k$ of them are heads and $n-k$ of them are
tails.  The probability of any particular sequence is $p^k(1-p)^{n-k}$.
For a sequence of $n$ coins, there are ${n \choose k}$ ways 
to pick a sequence with $k$ heads.  Thus, the probability
of exactly $k$ heads is 
\[
{n \choose k} p^k (1-p)^{n-k}
\]

\problempart
Consider the ratio of the probability of $k$ heads over the
probability of $k-1$ heads.
\begin{eqnarray*}
\frac{{n \choose k}p^k(1-p)^{n-k}}{{n \choose {k-1}}p^{k-1}(1-p)^{n-k+1}}
& = &\frac{\frac{n!}{k!(n-k)!}p}{\frac{n!}{(k-1)!(n-k+1)!(1-p)}} \\
& = & \frac{(n-k+1)p}{k(1-p)}
\end{eqnarray*}

This quantity is greater than 1 if and only if $(n-k+1)p>k(1-p)
\Longleftrightarrow np+p>k$.
So, for $k<np+p$ the probability increases from $k-1$ to $k$,
and for $k>np+p$, the probability decreases.
The value of $k$ that maximizes the probability is thus
$\lfloor np+p \rfloor$.

\problempart
The probability of no heads is $(1-p)^n= (1-\frac{1}{n})^n$.  For
large $n$, this expression approaches $1/e$.  

The probability of
exactly one head is ${n \choose 1}p(1-p)^{n-1} =
(1-\frac{1}{n})^{n-1}$.  Forming the ratio, we find the probability of
no heads over the probability of exactly one head is
$\frac{1}{1-1/n}$, which is one over the probability of a tail.

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\begin{eqnarray*}
Pr(A\cap \overline{B}) &=& Pr(A) - Pr(A\cap B) \\
     &=& Pr(A) - Pr(A) Pr(B) \hspace{0.5in}\mbox{since $A$ and $B$ are independent}\\
     &=& Pr(A)\cdot (1- Pr(B))\\
     &=& Pr(A)\cdot Pr(\overline{B})
\end{eqnarray*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\begin{eqnarray*}
Pr(R^2=r \cap e^T=t) &=& Pr[(R=\sqrt{r} \cup R=-\sqrt{r})\cap(e^T=t)]\\
&=& Pr[(R=\sqrt{r} \cup R=-\sqrt{r})\cap (T=\ln t)]\\
&=& Pr[((R=\sqrt{r}) \cap (T=\ln t)) \cup((R=-\sqrt{r})\cap (T=\ln t))]\\
&=& Pr[(R=\sqrt{r})\cap (T=\ln t)]+ Pr[(R=-\sqrt{r})\cap (T=\ln t)]\\
&=& Pr(R=\sqrt{r})Pr(T=\ln t)+Pr(R=-\sqrt{r})Pr(T=\ln t)~~~\mbox{since $R\& T$ ind.}\\
&=& Pr(R^2=r)Pr(e^T=t)
\end{eqnarray*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\bparts
\ppart
$Pr[\geq 0.01]=1-Pr[< 0.01] \leq 1 - ({{1-0.01}\over{1-0.02}})
2^{-2000}{2000\choose{0.01\times 2000}}$
\ppart
$Pr[\geq 600] = 1-Pr[< 600](n=1000,\alpha=3/5)\leq 1 - 
2^{-1000}{1000\choose {600}}{{1-3/5}\over{1-6/5}}$
\ppart
$Pr[\geq 200] = 1-Pr[< 200](n=1000,\alpha=1/5)\leq 1 - 
2^{-1000}{1000\choose {200}}{{1-1/5}\over{1-2/5}}$
\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \mbox{}

\begin{eqnarray*}
Pr(M_1=x) &=& Pr(max(D_1,D_2)=x)\\
&=& Pr(D_1 = x~ \mbox{and}~ D_2 < x) + Pr(D_1 < x~ \mbox{and}~ D_2 = x) +
Pr(D_1=D_2=x)\\
&=& {1\over 6}\cdot {{x-1\over 6}} + {x-1\over 6}\cdot {1\over 6} + {1\over 6}\cdot {1\over 6}\\
&=& {{2x-1}\over 36}
\end{eqnarray*}

$M_1$ and $M_2$ are not independent since we have:
\begin{eqnarray*}
Pr(M_1 = x | M_2 > x) &=& 0 ~~~~~~\mbox{for $x\in \{1,2,\ldots 6\}$} \\
                      &\not=& Pr(M_1 = x)
\end{eqnarray*}

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


\end{problems}

\end{document}
