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diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/prob-probspaces2.tex b/Master/texmf-dist/doc/latex/probsoln/samples/prob-probspaces2.tex new file mode 100644 index 00000000000..dacddb510cf --- /dev/null +++ b/Master/texmf-dist/doc/latex/probsoln/samples/prob-probspaces2.tex @@ -0,0 +1,43 @@ + % 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} |