\problemdata      
        {probability-intro-4} % latex-friendly label for the prob.
        {inclusion-exclusion} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 cp9W} % Usage (list ones you are aware of).
        {Albert Meyer, S02} % Last revision info (author, date).

\begin{problem}
Smith and Wesson are shooting at a target.  Suppose Smith's chance of
hitting the target is double that of his friend Wesson, and the
probability that at least one of them hits the target is $1/2$.
What is the probability that Wesson hits the target?  (Assume 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} \label{f1}\\
\pr{S \cup W}   & = &   \frac{1}{2}\label{f2}
\end{eqnarray}

Expanding the second equation using inclusion-exclusion gives:
\begin{align}
\frac{1}{2} & =  \pr{S \cup W} & \text{by~(\ref{f2})}\\
& = \pr{S} + \pr{W} - \pr{S \cap W} & \text{(inclusion-exclusion)}\\
& = \pr{S} + \pr{W} - \pr{S} \cdot \pr{W}   & \text{by independence}\\
& = 2 \pr{W} + \pr{W} - 2 \pr{W} \cdot \pr{W} & \text{by~(\ref{f1})}
\end{align}
Hence,
\begin{equation}\label{quad}
3 \pr{W} - 2 \pr{W}^2 - \frac{1}{2} = 0.
\end{equation}
Equation~(\ref{quad}) is a quadratic in $\pr{W}$; only one of its roots
makes lies between 0 and 1 and so makes sense as a probability, namely,
\[
\frac{3}{4} - \frac{1}{4} \sqrt{5} \approx 0.19.
\]}

\end{problem}
