\problemdata                                                             
        {probability-intro-3} % latex-friendly label for the prob.
        {Basic Probability} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 CP9M} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
Each morning Theory Pig pops out of his desk
drawer home--- all bright and perky--- and goes bounding off to play
around the Laboratory for Computer Science.  The desk drawer is
located at position $(0, 0)$.  At each time step, Theory Pig is
equally likely to bound north or east.  A northward bound increments
the first coordinate of his position, and an eastward bound increments
the second coordinate of his position.  (Assume that his direction at
one time step is independent of his direction at all other time
steps.)

At position $(4, 4)$, 6.042 TA Brian Dean is waiting with a bat, for
some reason.  What is the probability that Theory Pig visits position
$(4, 4)$ within his first 10 bounds?

\solution{ Each possible route can be described by the length 10 sequence
of \emph{N}'s and \emph{E}'s indicating the direction of a move; we use
these as the outcomes.  There are $2^{10}$ such outcomes.  Since \emph{N}
and \emph{E} equally likely, we assign uniform probabilities to the
outcomes.

Theory Pig visits position $(4, 4)$ if 4 of his first 8 bounds are north,
4 of his first 8 bounds are east, and his remaining two bounds are in any
direction.  Therefore, Theory Pig visits $(4, 4)$ in
\begin{equation*}
\binom{8}{4} \cdot 2^2   =    280
\end{equation*}
of these outcomes.  So the probability that Theory Pig visits
position $(4, 4)$ within his first 10 bounds is
\begin{equation*}
\frac{280}{2^{10}}  =  \frac{35}{128} \approx 0.27.
\end{equation*}
}
\end{problem}
                                 