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\begin{document}

Name: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Directions: {\bf \Large Work only on this sheet} (on both sides, if
needed).  MAKE SURE TO COPY YOUR ANSWERS TO A SEPARATE SHEET FOR SENDING
ME AN ELECTRONIC COPY LATER.

{\bf 1.} (20) The formal term used when two events cannot occur ``in the
same notebook line'' is that they are
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

{\bf 2.} Consider the simple board game in Section 2.10.  Let X denote
your position after your first turn.

\begin{itemize}

\item [(a)] (20) Find P(X = 1).

\item [(b)] (20) Find P(R = 1 $|$ X = 1).

\end{itemize}

{\bf 3.} Consider the ALOHA example, same as in the book, except that
both nodes start out inactive, i.e. $X_0 = 0$.  Assume p = 0.6, q = 0.2.

\begin{itemize}

\item [(a)] (20) Find the probability that there is a collision in first
epoch.

\item [(b)] (20) Find $P(X_1 = 2)$.

\end{itemize}

{\bf Solutions:}

{\bf 1.} disjoint

{\bf 2.a}

$$
P(X = 1) = P(R = 1 \textrm{ or }  R = 3 \textrm{ and }  B = 5)
= \frac{1}{6} + 
\frac{1}{6} \cdot
\frac{1}{6} 
$$

{\bf 2.b}

\begin{eqnarray}
P(R = 1 | X = 1) &=& 
\frac
{ P(R = 1 \textrm{ and } X = 1) }
{P(X = 1)} \\ 
&=& 
\frac
{ P(R = 1) }
{P(X = 1)} \\ 
&=& 
\frac
{
\frac{1}{6} 
}
{
\frac{1}{6} + 
\frac{1}{6} \cdot
\frac{1}{6} 
}
\end{eqnarray}

{\bf 3.a}  How can it happen?  A collision will occur in the
first epoch if and only both nodes develop messages and both try to
send, which has probability 

$$
q^2 p^2 = (0.2)^2 (0.6)^2
$$

{\bf 3.b}  How can it happen?  $X_1 = 2$ will occur if and only if both
nodes go active and either both send or both refrain from sending.  the
probability of that is

$$
q^2 [p^2 + (1-p)^2] = (0.2)^2 [0.6^2+0.4^2]
$$

\end{document}