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\title{Chapter 1}

\author{}

\date{}

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\textbf{1.} Rather complicated problem, but not bad if one is
persistent.  The probability in question is $P(L_2 = 1 | A_1 = 0$, where
$A_j$ is the number who alight at stop j.  One must then evaluate
$P(L_2 = 1 \textrm{ and } A_1 = 0)$ and $P(A_1 = 0$, by enumerating the
various probabilities.

\textbf{2.} Straightforward appilication of the mailing tubes.

\textbf{3.} Straightforward appilication of the mailing tubes.

\textbf{4.} No need for the mailing tubes here.  To find $P(S = 12 | T =
2)$, note that $T = 2$ means one of the pairs (1,1), (2,2),...,(6,6).
All are equally likely, so the conditional probability is 1/6.  The
opposite probability is 1.

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