

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Fall 1998 Final exam
% Added by George
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\problem {\bf (10 points)} Smith and Wesson are shooting at a target.
Smith's chance of hitting the target is double that of his friend
Wesson.  If they each take a shot, then the probability that at least
one of them hits the target is $\frac{1}{2}$.  What is the probability
that Wesson hits the target with a single shot?

(The event that one hits the target is independent of the event that
the other hits the target.)

\solution{Let $S$ be the event that Smith hits, and let $W$ be the
even that Wesson hits.  We are given two facts:

\begin{eqnarray*}
\Pr(S)      & = &   2 \Pr(W) \\
\Pr(S \cup W)   & = &   \frac{1}{2}
\end{eqnarray*}

Expanding the second equation using inclusion-exclusion gives:

\begin{eqnarray*}
\Pr(S) + \Pr(W) - \Pr(S \cap W) & = &   \frac{1}{2} \\
\Pr(S) + \Pr(W) - \Pr(S) \cdot \Pr(W)   & = &   \frac{1}{2} \\
2 \Pr(W) + \Pr(W) - 2 \Pr(W) \cdot \Pr(W) & = & \frac{1}{2}
\end{eqnarray*}

The second equation follows from independence, and the third equation
follows by substituting in the first fact listed above.  The root of
this quadratic that makes sense as a probabilty is $\frac{3}{4} -
\frac{1}{4} \sqrt{5} = 0.19\ldots$.  }