 % This file is public domain
 %
 % Finite probability spaces
\begin{defproblem}{weightedcoin}
\begin{onlyproblem}%
A coin is weighted so that heads is four times as likely
as tails. Find the probability that:
\begin{textenum}
\item tails appears,
\item heads appears
\end{textenum}
\end{onlyproblem}
\begin{onlysolution}%
Let $p=P(T)$, then $P(H)=4p$. We require $P(H)+P(T)=1$,
so $4p+p=1$, hence $p=\frac{1}{5}$. Therefore:
\begin{textenum}
\item $P(T)=\frac{1}{5}$,
\item $P(H)=\frac{4}{5}$
\end{textenum}
\end{onlysolution}
\end{defproblem}

\begin{defproblem}{validprobspaces}
\begin{onlyproblem}%
Under which of the following functions does 
$S=\{a_1,a_2\}$ become a probability space?
\par
\begin{textenum}
\begin{tabular}{ll}
\item $P(a_1)=\frac{1}{3}$, $P(a_2)=\frac{1}{2}$
&
\item\label{validprobspacescorrect1} $P(a_1)=\frac{3}{4}$, $P(a_2)=\frac{1}{4}$
\\
\item\label{validprobspacescorrect2} $P(a_1)=1$, $P(a_2)=0$
&
\item $P(a_1)=\frac{5}{4}$, $P(a_2)=-\frac{1}{4}$
\end{tabular}
\end{textenum}
\end{onlyproblem}%
\begin{onlysolution}%
\ref{validprobspacescorrect1} and \ref{validprobspacescorrect2}%
\end{onlysolution}
\end{defproblem}
