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\begin{document}

\psol{10}{November 22, 1996}

\bigskip

\begin{problems}

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\problem \mbox{}

The probability of winning the game remains the same but now there is
more than one optimal strategy.  If Carole reveals the ``Nerd Pride
Button'' it means that you must have chosen the box covering the
``Nerd Pocket Protector''.  In this case, you win with probability 1
if you switch. If Carole shows you the ``Nerd Pocket Protector'' then
there is $1/2$ probability that you have picked the prize box, so you
win with probability $1/2$ whether you switch or not.  Let NPB be the event
that Carole shows the ``Nerd Pride Button'', and NPP be the event that
Carole shows the ``Nerd Pocket Protector''. Then the 
total probability of winning is:
\begin{eqnarray*}
Pr(win) &=& Pr(win \mid \mbox{NPB})Pr(\mbox{NPB}) + Pr(win \mid\mbox{NPP})
Pr(\mbox{NPP}) \\
        &=& 1 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} \\
        &=& \frac{2}{3}
\end{eqnarray*}

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\problem \mbox{}

\bparts
\ppart
In this case, you should always pick Box 1.  Monty (Prof. Leighton)
will be forced to reveal the location of the other ``Nerd Pride'' pin.
Upon switching, you will be guaranteed to win the ice cream (i.e.
the probability of winning is 1).  If you
pick a box besides 1, Monty can always reveal Box 1 (giving you no
additional information), and you will have a 1/2 chance of winning.

\ppart
If you play the ``don't-switch strategy'', you will win whenever you
pick the correct box initially.  Thus, you will win with probability
1/3.  If you play the ``switch strategy'', you will lose every time
you initially pick the wrong box, since Monty can decide to end the game.
Everytime you pick the right box, Monty will reveal a bad box, and
then you will switch to a bad box.  Thus, you lose again.
Therefore, the probability of winning is 0 with the ``switch strategy''.
\eparts

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\problem

Let $A_i$ be the event that componet $i$ fails. Let $A$ be the probability
that one or more components fail. Then we have:
\begin{eqnarray*}
Pr(A) &=& \sum_{i=1}^{1000}{Pr(A_i)}\\
  &=& 1000\cdot Pr(A_i)\\
  &\leq& 1000\cdot \frac{1}{10,000}~~~~~~\mbox{since $Pr(A_i)\leq 1/10,000$}\\
  &=& 1/10
\end{eqnarray*}

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\problem

Let $B$ be the event the criminal has a beard.  Let $W$ be the
event the witness says the criminal has a beard.
We want to compute $Pr(B|W)$.  By Bayes's theorem,
\[
Pr(B|W)=\frac{Pr(W|B)Pr(B)}{Pr(W|B)Pr(B)+Pr(W|\overline{B})Pr(\overline{B})}
= \frac{(0.8)(.15)}{(0.8)(0.15)+(0.12)(0.85)} =0.54
\]

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\problem

\bparts
\ppart 
\begin{eqnarray*}
1 - Pr(A|C) &=& 1 - {{Pr(A\cap C)}\over {Pr(C)}} \\
            &=& {{Pr(C) - Pr(A\cap C)}\over {Pr(C)}} \\
            &=& {{Pr(\overline{A}\cap C)}\over {Pr(C)}}\\
            &=& Pr(\overline{A}|C)
\end{eqnarray*}
\ppart

Yes, it is.  
Consider the uniform sample space $S=\{1,2,\ldots, 2 \cdot 10^6\}$. \\
Let $A = \{1\}, B = \{1,2,\ldots,10^6 \}, C = \{1,10^6+1,\ldots,2 \cdot 10^6 -1\}$. \\
Then $Pr(A \mid  B)  = Pr(A \mid C) = 10^{-6}$, while $Pr(A \mid B \cap C) = 1$.
 
\eparts

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\end{problems}

\end{document}










