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author | Norbert Preining <norbert@preining.info> | 2019-09-02 13:46:59 +0900 |
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committer | Norbert Preining <norbert@preining.info> | 2019-09-02 13:46:59 +0900 |
commit | e0c6872cf40896c7be36b11dcc744620f10adf1d (patch) | |
tree | 60335e10d2f4354b0674ec22d7b53f0f8abee672 /macros/latex/contrib/probsoln/samples/prob-probspaces.tex |
Initial commit
Diffstat (limited to 'macros/latex/contrib/probsoln/samples/prob-probspaces.tex')
-rw-r--r-- | macros/latex/contrib/probsoln/samples/prob-probspaces.tex | 33 |
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diff --git a/macros/latex/contrib/probsoln/samples/prob-probspaces.tex b/macros/latex/contrib/probsoln/samples/prob-probspaces.tex new file mode 100644 index 0000000000..4862e5caef --- /dev/null +++ b/macros/latex/contrib/probsoln/samples/prob-probspaces.tex @@ -0,0 +1,33 @@ + % This file is public domain + % + % Finite probability spaces +\newproblem{weightedcoin}{% +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}}{% +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}} + +\newproblem*{validprobspaces}{% +Under which of the following functions does +$S=\{a_1,a_2\}$ become a probability space? +\par +\begin{textenum} +\begin{tabular}{ll} +\incorrectitem $P(a_1)=\frac{1}{3}$, $P(a_2)=\frac{1}{2}$ +& +\correctitem $P(a_1)=\frac{3}{4}$, $P(a_2)=\frac{1}{4}$ +\\ +\correctitem $P(a_1)=1$, $P(a_2)=0$ +& +\incorrectitem $P(a_1)=\frac{5}{4}$, $P(a_2)=-\frac{1}{4}$ +\end{tabular} +\end{textenum} +} |