% 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}