\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Conditional} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS10; F00} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
In London, half of the days have some rain. The weather forecaster is
correct 2/3 of the time -- i.e. the probability that it rains, given
that she has predicted that it will rain, and the probability that it
does not rain given that she has predicted that it does not rain, are
both equal to 2/3. When rain is forecast, Mr.Paddington takes his
umbrella. When rain is not forecast, he takes it with probability
1/3.
\bparts

\ppart Find the probability of the weather forecaster forecasting rain.
\solution{

\textbf{Solution:}
To compute this probability, first we define the following events:

\begin{eqnarray*}
F & \eqdef & \text{the weather forecaster forecasts rain},\\
R & \eqdef & \text{there is rain in London} ,\\
U & \eqdef & \text{Mr.Paddington takes his umbrella  }.
\end{eqnarray*}

Using the law of total probability, we have

\[
\prob{R} = \prob{R \mid F} \prob{F} + \prob{R \mid {\overline F}} \prob{{\overline F}}
\]

Plugging in values we get:
\begin{eqnarray*}
\frac{1}{2} &= & \frac{2}{3}\pr{F} +\frac{1}{3}\pr{{\overline F}}\\
&= & \frac{2}{3}\pr{F} + \frac{1}{3}(1-\pr{F}) \\
&= & \frac{1}{3}\pr{F} + \frac{1}{3}
\end{eqnarray*}

Therefore, $\pr{F} = \frac{1}{2}$.
}

\ppart Find the probability that Paddington has no umbrella, given that it rains.

\solution{We construct a tree as shown below. 

\begin{figure}[htb]
\begin{center}
\includegraphics[width=4in]{figures/Probability_Tree_Figure}
\end{center}
\caption{Probability Tree}
\label{fig:Probability_Tree_Figure}
\end{figure}


We wish to find
\begin{eqnarray*}
\prob{\text{no umbrella}  \mid \text{rain}}
  & = &\frac{\prob{\mbox{no umbrella} \wedge \mbox{rain}}}{\prob{\mbox{rain}}}\\
& = & \frac{1/9}{1/2} = 2/9.
\end{eqnarray*}
}

\ppart Find the probability that it doesn't rain, given that he brings his
umbrella.

\solution{
we wish to find
\begin{eqnarray*}
\prob{\text{no rain} \mid \text{umbrella}} & = &
 \frac{\prob{\mbox{no rain} \wedge \mbox{umbrella}}}{\prob{\mbox{umbrella}}}\\
& = & \frac{1/6 + 1/9}{1/9 + 1/6 + 1/3 + 1/18} = 5/12
\end{eqnarray*}
}

\eparts

\end{problem}