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- % 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}