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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\begin{document}


\section{Exam}

\subsubsection{Comment}

I detta exempel testar vi studentens kunskaper i grundl\"{a}ggande
sannolikhetsl\"{a}ra.

\subsubsection{Text}

\section{Sannolikhetsl\"{a}ra}

Eftersom uppgifterna r\"{a}ttas av en dator \"{a}r det viktigt att du
skriver p\aa\ samma s\"{a}tt som i boken.

Vid betingning anv\"{a}nds symbolen $\mid $ och den erh\aa lls via CTRL+mid.

\subsection{Comment}

seed:=12345

\subsection{Setup}

Errors: report

Choices: No Break

Title: Probability

Submit:Click to Grade

$\limfunc{nplaces}(x,n)=1.0\left\lfloor 10^{n}x+0.5\right\rfloor /10^{n}$

\section{Part}

\section{Text}

\section{Teoridel}

\section{Question}

\subsection{Setup}

Select: 5

\subsection{Variant}

\subsubsection{Setup}

Choices: Check, Break

\subsubsection{Statement}

Vilka av f\"{o}ljande p\aa st\aa enden \"{a}r en del av Komogorovs
axiomsystem?

\subsubsection{Choices}

\begin{itemize}
\item[1] $P\left( \Omega \right) =1$

\item[-1] F\"{o}r alla delm\"{a}ngder g\"{a}ller att deras sannolikhet
ligger mellan $0$ och $1$.

\item[1] Om sannolikheter f\"{o}r tv\aa\ disjunkta delm\"{a}ngder adderas s%
\aa\ erh\aa lls sannolikheten f\"{o}r det gemensamma.
\end{itemize}

\subsubsection{Solution}

\begin{enumerate}
\item \"{A}r en del av axiomsystemet.

\item Sannolikheter kan anta v\"{a}rdena $0$ och $1$, som texten \"{a}r
formulerad ing\aa r inte dessa v\"{a}rden.

\item \"{A}r en del av axiomsystemet.
\end{enumerate}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\cup B\right) =P\left( A\right) +P\left( B\right)
-P\left( A\cap B\right) $

Points: 3

\subsubsection{Statement}

Ange formeln f\"{o}r de allm\"{a}nna additionssatsen.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) $

Points: 3

\subsubsection{Statement}

Ange formeln f\"{o}r den speciella additionssatsen.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\cap B\right) =P\left( A\mid B\right) P\left(
B\right) $

Points: 3

\subsubsection{Statement}

Ange formeln f\"{o}r den allm\"{a}na multiplikationssatsen.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\cap B\right) =P\left( A\right) P\left( B\right) $

Points: 3

\subsubsection{Statement}

Ange formeln f\"{o}r den speciella multiplikationssatsen.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( \complement A\right) =1-P\left( A\right) $

Points: 3

\subsubsection{Statement}

Ange formeln f\"{o}r komplementsatsen.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A_{i}\mid A\right) =\frac{P\left( A\mid
A_{i}\right) P\left( A_{i}\right) }{\sum_{i=1}^{n}P\left( A\mid A_{i}\right)
P\left( A_{i}\right) }$

Points: 3

\subsubsection{Statement}

Ange formeln i Bayes sats.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=p=\frac{g}{m}$

Points: 3

\subsubsection{Statement}

Hur ser den klassiska sannolikhetsdefinitionen ut (anv\"{a}nd bokst\"{a}%
verna $p$, $g$ och $m$)?

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\mid B\right) =P\left( A\right) $

Points: 3

\subsubsection{Statement}

Ange den "naturliga" definitionen av oberoende f\"{o}r tv\aa\ m\"{a}ngder $A$
och $B$.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\cap B\right) =P\left( A\right) P\left( B\right) $

Points: 3

\subsubsection{Statement}

Ange den utvidgningsbara definitionen av oberoende f\"{o}r tv\aa\ m\"{a}%
ngder $A$ och $B$.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=P\left( A\mid B\right) =\frac{P\left( A\cap B\right) }{%
P\left( B\right) }$

Points: 3

\subsubsection{Statement}

Hur lyder definitionen av betingning f\"{o}r tv\aa\ m\"{a}ngder $A$ och $B$.

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\sum_{i=1}^{n}P\left( A_{i}\right) $

Points: 3

\subsubsection{Statement}

Vad \"{a}r sannolikheten f\"{o}r h\"{a}ndelsen $A_{1}\cup A_{2}\cup \cdots
\cup A_{n}$ n\"{a}r alla ing\aa ende h\"{a}ndelser \"{a}r parvis disjunkta?

\subsubsection{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsection{Variant}

\subsubsection{Setup}

Choices: Check, Break

\subsubsection{Statement}

Bocka f\"{o}r de p\aa st\aa enden som \"{a}r korrekta.

\subsubsection{Choices}

\begin{itemize}
\item[-2] Med unionen av tv\aa\ m\"{a}ngder menas det som \"{a}r gemensamt f%
\"{o}r dem.

\item[1] Med snittet av tv\aa\ m\"{a}ngder menas det som \"{a}r gemensamt f%
\"{o}r dem.

\item[1] Differensen, $A\setminus B$, mellan tv\aa\ m\"{a}ngder $A$ och $B$ 
\"{a}r det som finns i $A$ men inte i $B$.

\item[-2] Tv\aa\ m\"{a}ngder som \"{a}r disjunkta \"{a}r ocks\aa\ oberoende.

\item[1] M\"{a}ngden av alla m\"{o}jliga utfall kallas utfallsrum.

\item[-2] Sannolikheten f\"{o}r den tomma m\"{a}ngden \"{a}r st\"{o}rre \"{a}%
n noll.

\item[1] Det finns fyra m\"{a}ngder som \"{a}r parvis oberoende men \"{a}%
ndock ej oberoende?
\end{itemize}

\section{Part}

\section{Text}

\section{Problemdel}

\section{Question}

\subsection{Comment}

Klassisk sannolikhetsdefinition

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{5}{36},\frac{1}{6},\frac{1}{36},\frac{5}{18},%
\frac{1}{9},\frac{5}{36}\right) $

\subsubsection{Statement}

Tv\aa\ symmetriska sexsidiga t\"{a}rningar kastas och deras summa ber\"{a}%
knas. S\"{a}tt 
\begin{align*}
A& =\text{ summan \"{a}r }8 \\
B& =\text{de tv\aa\ t\"{a}rningarna visar samma resultat}
\end{align*}%
Ber\"{a}kna f\"{o}ljande sannolikheter (svara p\aa\ formen $\frac{p}{q}$):

\paragraph{Substatement}

$P\left( A\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\paragraph{Substatement}

$P\left( B\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Substatement}

$P\left( A\cup B\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\paragraph{Substatement}

$P\left( A\cap B\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{4}$)

\paragraph{Substatement}

$P\left( A\setminus B\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{5}$)

\paragraph{Substatement}

$P\left( B\setminus A\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{6}$)

\subsubsection{Solution}

Av figur f\"{o}ljer att%
\begin{eqnarray*}
P\left( A\right) &=&\frac{5}{36} \\
P\left( B\right) &=&\frac{6}{36} \\
P\left( A\cap B\right) &=&\frac{1}{36}
\end{eqnarray*}%
detta ger oss%
\begin{eqnarray*}
P\left( A\cup B\right) &=&P\left( A\right) +P\left( B\right) -P\left( A\cap
B\right) \\
&=&\frac{5}{36}+\frac{6}{36}-\frac{1}{36}=\frac{10}{36} \\
P\left( A\setminus B\right) &=&P\left( A\right) -P\left( A\cap B\right) \\
&=&\frac{5}{36}-\frac{1}{36}=\frac{4}{36} \\
P\left( B\setminus A\right) &=&P\left( B\right) -P\left( A\cap B\right) \\
&=&\frac{6}{36}-\frac{1}{36}=\frac{5}{36}
\end{eqnarray*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{1}{16},\frac{1}{4},\frac{3}{8},\frac{1}{4},%
\frac{1}{16}\right) $

\subsubsection{Statement}

Ett symmetrisk mynt kastas 4 g\aa nger. Best\"{a}m sannolikheterna f\"{o}r
att erh\aa lla $0$, $1$, $2$, $3$ och $4$ krona (svara p\aa\ formen $\frac{p%
}{q}$)

\paragraph{Substatement}

$P\left( \text{noll krona}\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\paragraph{Substatement}

$P\left( \text{en krona}\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Substatement}

$P\left( \text{tv\aa\ krona}\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\paragraph{Substatement}

$P\left( \text{tre krona}\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{4}$)

\paragraph{Substatement}

$P\left( \text{fyra krona}\right) $

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{5}$)

\subsubsection{Solution}

Ett mynt kan visa antingen krona eller klave och om vi betecknar mynt $k$:s
resultat med $A_{k}$ kan den beskrivna situationen skrivas som $\left(
A_{1},A_{2},A_{3},A_{4}\right) $. Det totala antalet m\"{o}jliga utfall blir
d\aa\ enligt multiplikationsprincipen $2\times 2\times 2\times 2=2^{4}$. De
gynnsamma utfallen f\"{o}r att f\aa\ $0$ krona erh\aa lls d\aa\ alla $A_{i}$
blir klave dvs det finns ett s\aa dant fall -- $\left( kl,kl,kl,kl\right) $.
De gynnsamma utfallen f\"{o}r att f\aa\ $1$ krona erh\aa lls d\aa\ alla utom
ett $A_{i}$ blir klave och dessa \"{a}r $\left( kr,kl,kl,kl\right) $, $%
\left( kl,kr,kl,kl\right) $, $\left( kl,kl,kr,kl\right) $ och $\left(
kl,kl,kl,kr\right) $ dvs $4$ gynnsamma utfall. P\aa\ samma s\"{a}tt erh\aa %
ller vi att de gynnsamma utfallen f\"{o}r $2$ krona \"{a}r $6$, f\"{o}r tre
krona \"{a}r $4$ och f\"{o}r $4$ krona $1$. De s\"{o}kta sannolikheterna kan
d\"{a}rf\"{o}r skrivas%
\begin{align*}
P\left( 0\text{ krona}\right) & =\frac{1}{16} \\
P\left( 1\text{ krona}\right) & =\frac{4}{16} \\
P\left( 2\text{ krona}\right) & =\frac{6}{16} \\
P\left( 3\text{ krona}\right) & =\frac{4}{16} \\
P\left( 4\text{ krona}\right) & =\frac{1}{16}
\end{align*}%
Ett alternativt s\"{a}tt att finna de gynnsamma utfallen \"{a}r att utnyttja
den diskreta matematiken som s\"{a}ger att antalet m\"{o}jligheter att ta $k$
element ur $4$ \"{a}r $\binom{4}{k}$ och d\"{a}refter konstatera att%
\begin{equation*}
P\left( k\text{ krona}\right) =\frac{\binom{4}{k}}{2^{4}}
\end{equation*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{3}{4}$

\subsubsection{Statement}

Vid kast med tv\aa\ symmetriska mynt erh\aa lls utfallsrummet 
\begin{equation*}
\Omega =\left\{ \left( G,G\right) ,\left( G,K\right) ,\left( K,G\right)
,\left( K,K\right) \right\}
\end{equation*}%
d\"{a}r G=gubbe och K=krona. F\"{o}r var och en av de m\"{o}jliga paren erh%
\aa lls d\"{a}rf\"{o}r sannolikheten $\frac{1}{4}$. S\"{a}tt 
\begin{align*}
E& =\left\{ \left( G,G\right) ,\left( G,K\right) \right\} \\
F& =\left\{ \left( G,G\right) ,\left( K,G\right) \right\}
\end{align*}%
Best\"{a}m $P\!\left( E\cup F\right) $ (svara p\aa\ formen $\frac{p}{q}$).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsubsection{Solution}

Det finns tv\aa\ s\"{a}tt att resonera

\begin{enumerate}
\item $P\left( E\cup F\right) =P\left( E\right) +P\left( F\right) -P\left(
E\cap F\right) =\frac{2}{4}+\frac{2}{4}-\frac{1}{4}$

\item $E\cup F=\left\{ \left( G,G\right) ,\left( G,K\right) ,\left(
K,G\right) \right\} $ varav $P\left( E\cup F\right) =\frac{3}{4}$
\end{enumerate}

\section{Question}

\subsection{Comment}

Kolmogorovs axiom

\subsection{Variant}

\subsubsection{Setup}

Choices: permute

$p:=\limfunc{nplaces}\left( \func{rand}\left( 0,1000\right) /1000,1\right) $

$q:=\limfunc{nplaces}\left( \func{rand}\left( 0,1000\right) /1000,1\right) $

$I:=\left\{ 
\begin{array}{ccc}
1 & if & p+q\leq 1 \\ 
2 & if & p+q>1%
\end{array}%
\right. $

$\limfunc{svar}:=(\func{ja},\func{nej})$

\subsubsection{Statement}

Tv\aa\ h\"{a}ndelser $A$ och $B$ har sannolikheterna $%
%TCIMACRO{\FORMULA{p}{p}{evaluatenum}}%
%BeginExpansion
p%
%EndExpansion
$ respektive $%
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
$ att intr\"{a}ffa. Kan h\"{a}ndelserna vara disjunkta?

\paragraph{Choices}

\begin{itemize}
\item $%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{(I\func{mod}2)+1}}{\left( \limfunc{svar}\right) \left[ I+1\right] }{evaluate}}%
%BeginExpansion
\left( \limfunc{svar}\right) \left[ I+1\right] %
%EndExpansion
$

\item $%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{I}}{\limfunc{svar}\left( I\right) }{evaluate}}%
%BeginExpansion
\limfunc{svar}\left( I\right) %
%EndExpansion
$\correctchoice{}
\end{itemize}

\subsubsection{Solution}

Det g\"{a}ller att 
\begin{equation*}
P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) -P\left( A\cap
B\right)
\end{equation*}%
och om h\"{a}ndelserna \"{a}r disjunkta g\"{a}ller \"{a}ven 
\begin{equation*}
P\left( A\cap B\right) =\varnothing
\end{equation*}%
H\"{a}rav f\"{o}ljer att 
\begin{equation*}
P\left( A\cup B\right) =%
%TCIMACRO{\FORMULA{p+q}{p+q}{evaluate}}%
%BeginExpansion
p+q%
%EndExpansion
\end{equation*}%
vilket \"{a}r om\"{o}jligt.

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{1}{4}$

\subsubsection{Statement}

Visa att sannolikheten att exakt en av h\"{a}ndelserna $A$ och $B$ ($%
A\triangle B$) intr\"{a}ffar \"{a}r 
\begin{equation*}
P\left( A\triangle B\right) =P\left( A\right) +P\left( B\right) -2P\left(
A\cap B\right)
\end{equation*}%
samt best\"{a}m denna sannolikhet n\"{a}r tv\aa\ symmetriska sexsidiga t\"{a}%
rningar kastas och 
\begin{align*}
A& =\text{ summan \"{a}r }8\text{,} \\
B& =\text{de tv\aa\ t\"{a}rningarna visar samma resultat.}
\end{align*}%
(svara p\aa\ formen $\frac{p}{q}$):

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsubsection{Solution}

H\"{a}ndelsen \textquotedblright exakt en av h\"{a}ndelserna $A$ och $B$%
\textquotedblright\ kan skrivas $\left( A\setminus B\right) \cup \left(
B\setminus A\right) $ varav f\"{o}ljer 
\begin{align*}
P\left( \left( A\setminus B\right) \cup \left( B\setminus A\right) \right) &
=P\left( A\setminus B\right) +P\left( B\setminus A\right) \\
& =P\left( A\right) -P\left( A\cap B\right) +P\left( B\right) -P\left( A\cap
B\right) \\
& =P\left( A\right) +P\left( B\right) -2P\left( A\cap B\right)
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

Choices: permute

\subsubsection{Statement}

Tv\aa\ disjunkta h\"{a}ndelser $A$ och $B$ har sannolikheter skilda fr\aa n
noll. Kan de vara oberoende?

\paragraph{Choices}

\begin{itemize}
\item Ja

\item Nej\correctchoice{}
\end{itemize}

\subsubsection{Solution}

Att h\"{a}ndelserna \"{a}r disjunkta medf\"{o}r att $P\left( A\cap B\right)
=0$ men eftersom sannolikheterna \"{a}r st\"{o}rre \"{a}n noll g\"{a}ller 
\begin{equation*}
P\left( A\right) P\left( B\right) >0
\end{equation*}%
och s\aa ledes kan det ej g\"{a}lla att $P\left( A\cap B\right) =P\left(
A\right) P\left( B\right) $.

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}\left( \func{rand}\left( 100,200\right)
/1000,2\right) $

$p_{2}:=\limfunc{nplaces}\left( \func{rand}\left( 50,100\right)
/1000,2\right) $

$p_{12}:=\limfunc{nplaces}\left( \func{rand}\left( 10,30\right)
/1000,2\right) $

$\limfunc{svar}:=\left( \limfunc{nplaces}\left( p_{1}+p_{2}-p_{12},2\right) ,%
\limfunc{nplaces}\left( 1-p_{1}-p_{2}+p_{12},2\right) ,\limfunc{nplaces}%
\left( p_{1}+p_{2}-2p_{12},2\right) \right) $

\subsubsection{Statement}

Vid tillverkning av apelsinf\"{a}rgade klinkers kan tv\aa\ fel uppst\aa\ vid
br\"{a}nningen: Fel av typ 1 best\aa r av att plattan f\aa r
\textquotedblright smutsfl\"{a}ckar\textquotedblright\ och detta intr\"{a}%
ffar med sannolikheten $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$. Fel av typ 2 \"{a}r sprickbildning som g\"{o}r att plattan m\aa ste
kasseras. Detta fel intr\"{a}ffar med sannolikheten $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$. Ibland, med sannolikhet $%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate}}%
%BeginExpansion
p_{12}%
%EndExpansion
$, f\aa r en platta fel av b\aa de typ 1 och 2. Best\"{a}m f\"{o}ljande
sannolikheter: En platta

\begin{enumerate}
\item f\aa r minst ett av felen

\item blir felfri

\item f\aa r exakt ett av felen
\end{enumerate}

(svara med 2 decimaler).

\paragraph{Substatement}

f\aa r minst ett av felen

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\paragraph{Substatement}

blir felfri

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Substatement}

f\aa r exakt ett av felen

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\subsubsection{Solution}

S\"{a}tt 
\begin{align*}
A& =\text{fel av typ 1} \\
B& =\text{fel av typ 2}
\end{align*}%
det g\"{a}ller d\aa\ 
\begin{align*}
P\left( A\right) & =%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate} }%
%BeginExpansion
p_{1}
%EndExpansion
\\
P\left( B\right) & =%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate} }%
%BeginExpansion
p_{2}
%EndExpansion
\\
P\left( A\cap B\right) & =%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate}}%
%BeginExpansion
p_{12}%
%EndExpansion
\end{align*}%
varvid

\begin{enumerate}
\item 
\begin{align*}
P\left( A\cup B\right) & =P\left( A\right) +P\left( B\right) -P\left( A\cap
B\right) \\
& =%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate} }%
%BeginExpansion
p_{12}
%EndExpansion
\\
& =%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{1}}{\left( \limfunc{svar}\right) 1}{evaluate}}%
%BeginExpansion
\left( \limfunc{svar}\right) 1%
%EndExpansion
\end{align*}

\item 
\begin{equation*}
1-P\left( A\cup B\right) =%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{2}}{\left( \limfunc{svar}\right) 2}{evaluate}}%
%BeginExpansion
\left( \limfunc{svar}\right) 2%
%EndExpansion
\end{equation*}

\item 
\begin{align*}
P\left( \left( A\setminus B\right) \cup \left( B\setminus A\right) \right) &
=P\left( A\right) -P\left( A\cap B\right) +P\left( B\right) -P\left( B\cap
A\right) \\
& =%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
-2%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate} }%
%BeginExpansion
p_{12}
%EndExpansion
\\
& =%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{3}}{\left( \limfunc{svar}\right) 3}{evaluate}}%
%BeginExpansion
\left( \limfunc{svar}\right) 3%
%EndExpansion
\end{align*}
\end{enumerate}

\section{Question}

\subsection{Comment}

Betingning

\subsection{Variant}

\subsubsection{Setup}

$n:=\func{rand}\left( 10,20\right) $

$k:=\func{rand}(4,n-5)$

$\limfunc{svar}:=\frac{1}{n-k+1}$

\subsubsection{Statement}

I en hatt finns $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ kort numrerade fr\aa n $1$ till $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ och ett av dessa dras. Om vi nu f\aa r den extra informationen att det
dragna kortets val\"{o}r \"{a}r \aa tminstone $%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
$ vad \"{a}r d\aa\ sannolikheten att det dragna kortet \"{a}r $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ (svara p\aa\ formen $\frac{p}{q}$)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt 
\begin{align*}
E& =\text{det dragna kortet \"{a}r }%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }%
%BeginExpansion
n
%EndExpansion
\\
F& =\text{kortets val\"{o}r \"{a}r \aa tminstone }%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
\end{align*}%
vi erh\aa ller d\aa 
\begin{align*}
P\left( E\mid F\right) & =\frac{P\left( E\cap F\right) }{P\left( F\right) }=%
\frac{P\left( E\right) }{P\left( F\right) }=\frac{\frac{1}{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}}{\frac{%
%TCIMACRO{\FORMULA{n-k+1}{n-k+1}{evaluate}}%
%BeginExpansion
n-k+1%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}} \\
& =\frac{1}{%
%TCIMACRO{\FORMULA{n-k+1}{n-k+1}{evaluate}}%
%BeginExpansion
n-k+1%
%EndExpansion
}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$n:=\func{rand}(2)$

$k:=\func{rand}(2)$

$\limfunc{svar}:=\frac{1}{3}$

\subsubsection{Statement}

En familj har $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$\ barn. Vad \"{a}r sannolikheten att b\aa da \"{a}r pojkar om \aa tminstone
en av dem \"{a}r en pojke? Ledning: Om $n=2$ s\aa\ \"{a}r ett l\"{a}mpligt
utfallsrum%
\begin{equation*}
\Omega =\left\{ \left( p,p\right) ,\left( p,f\right) ,\left( f,p\right)
,\left( f,f\right) \right\}
\end{equation*}%
d\"{a}r alla utfall \"{a}r lika sannolika (svara p\aa\ formen $\frac{p}{q}$).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt 
\begin{align*}
E& =%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
\text{ av barnen \"{a}r pojkar} \\
F& =\text{\aa tminstone ett av barnen \"{a}r en pojke}
\end{align*}%
varav s\"{o}kt sannolikhet erh\aa lls till 
\begin{align*}
P\left( E\mid F\right) & =\frac{P\left( E\cap F\right) }{P\left( F\right) }=%
\frac{P\left( \left\{ \left( p,p\right) \right\} \right) }{P\left( \left\{
\left( p,p\right) ,\left( p,f\right) ,\left( f,p\right) \right\} \right) } \\
& =\frac{\frac{1}{4}}{\frac{3}{4}} \\
& =\frac{1}{3}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}(\func{rand}(100,400)/1000,2)$

$p_{2}:=\limfunc{nplaces}(\func{rand}(400,600)/1000,2)$

$\limfunc{svar}:=p_{1}\times p_{2}$

\subsubsection{Statement}

Pia v\"{a}ljer mellan en kurs i objektorienterad programmering och en kurs i
teknikhistoria. Om hon v\"{a}ljer den f\"{o}rsta kursen har hon
sannolikheten $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$ att erh\aa lla h\"{o}gsta betyg och motsvarande sannolikhet f\"{o}r den
senare kursen \"{a}r en $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$. Pia:s valmetod \"{a}r slantsingling med ett symmetriskt mynt. Vad \"{a}r
sannolikheten att hon f\aa r h\"{o}gsta betyg i objektorienterad
programmering (svara med tv\aa\ decimaler)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt 
\begin{align*}
E& =\text{Pia v\"{a}ljer objektorienterad programmering} \\
F& =\text{Pia erh\aa ller h\"{o}gsta betyg}
\end{align*}%
d\aa\ \"{a}r s\"{o}kt sannolikhet 
\begin{align*}
P\left( E\cap F\right) & =P\left( F\mid E\right) P\left( E\right) \\
& =%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate} }%
%BeginExpansion
p_{2}
%EndExpansion
\\
& =%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{evaluate}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\end{align*}

\section{Question}

\subsection{Comment}

Oberoende

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{1}{8}$

\subsubsection{Statement}

Parvis oberoende \"{a}r inte detsamma som oberoende vilket inses av f\"{o}%
ljande: En urna inneh\aa ller $4$ bollar numrerade $1,2,3,4$. Tag en av
dessa och beteckna med $1$ att boll $1$ valts, med $2$ att boll $2$ valts
osv. S\"{a}tt sedan 
\begin{align*}
E& =\left\{ 1,2\right\} \\
F& =\left\{ 1,3\right\} \\
G& =\left\{ 1,4\right\}
\end{align*}%
Visa att dessa h\"{a}ndelser \"{a}r parvis oberoende men ej oberoende. Ange
vad $P\left( E\right) P\left( F\right) P\left( G\right) $ \"{a}r (svara p%
\aa\ formen $\frac{p}{q}$).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

Det g\"{a}ller att 
\begin{align*}
P\left( E\cap F\right) & =P\left( \left\{ 1\right\} \right) =\frac{1}{4} \\
P\left( E\right) P\left( F\right) & =\frac{2}{4}\frac{2}{4}=\frac{1}{4}
\end{align*}%
och s\aa ledes \"{a}r $E$ och $F$ oberoende h\"{a}ndelser. P\aa\ samma s\"{a}%
tt visas att 
\begin{align*}
P\left( E\cap G\right) & =P\left( E\right) P\left( G\right) =\frac{1}{4} \\
P\left( F\cap G\right) & =P\left( F\right) P\left( G\right) =\frac{1}{4}
\end{align*}%
Vidare \"{a}r 
\begin{align*}
P\left( E_{1}\cap E_{2}\cap E_{3}\right) & =\left\{ 1\right\} =\frac{1}{4} \\
P\left( E\right) P\left( F\right) P\left( G\right) & =\frac{1}{8}
\end{align*}%
men detta betyder att h\"{a}ndelserna f\"{o}rvisso \"{a}r parvis oberoende
men alla tillsammans \"{a}r beroende.

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{1}{8}$

\subsubsection{Statement}

Vid kast med tv\aa\ perfekt symmetriska t\"{a}rningar bildas h\"{a}ndelserna%
\begin{align*}
A& =\text{j\"{a}mnt antal prickar t\"{a}rning }1 \\
B& =\text{j\"{a}mnt antal prickar t\"{a}rning }2 \\
C& =\text{t\"{a}rningssumman \"{a}r }7
\end{align*}%
Visa att dessa h\"{a}ndelser \"{a}r parvis oberoende men ej oberoende.

\paragraph{Choices}

\paragraph{Solution}

F\"{o}ljande sannolikheter erh\aa lls%
\begin{align*}
P\left( A\right) & =\frac{3}{6} \\
P\left( B\right) & =\frac{3}{6} \\
P\left( C\right) & =\frac{6}{36}
\end{align*}%
vilket ger tabellen%
\begin{align*}
P\left( A\cap B\right) & =\frac{9}{36}=\frac{3}{6}\frac{3}{6}=P\left(
A\right) P\left( B\right) \\
P\left( A\cap C\right) & =\frac{3}{36}=\frac{3}{6}\frac{6}{36}=P\left(
A\right) P\left( C\right) \\
P\left( B\cap C\right) & =\frac{3}{36}=\frac{3}{6}\frac{6}{36}=P\left(
B\right) P\left( C\right)
\end{align*}%
varf\"{o}r h\"{a}ndelserna \"{a}r parvis oberoende. Men%
\begin{equation*}
P\left( A\cap B\cap C\right) =0\neq \frac{3}{6}\frac{3}{6}\frac{6}{36}%
=P\left( A\right) P\left( B\right) P\left( C\right)
\end{equation*}%
och d\"{a}rf\"{o}r \"{a}r de tre h\"{a}ndelserna inte oberoende.

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}(\func{rand}(900,900)/1000,2)$

$p_{2}:=\limfunc{nplaces}(\func{rand}(800,800)/1000,2)$

$\limfunc{svar}:=\limfunc{nplaces}\left( p_{1}^{3}+p_{2}^{2}-p_{1}^{3}\times
p_{2}^{2},2\right) $

\subsubsection{Statement}

Ett signalsystem best\aa r av fem komponenter som \"{a}r kopplade p\aa\ f%
\"{o}ljande s\"{a}tt: Komponenterna $1$, $2$ och $3$ \"{a}r liksom
komponenterna $4$ och $5$ kopplade i serie. De tv\aa\ serierna \"{a}r sedan
parallellkopplade. Antag att komponenterna $1$, $2$ och $3$ alla har
sannolikheten $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$ att fungera samt att komponenterna $4$ och $5$ har funktionssannlikheten $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$. Komponenterna fungerar oberoende av varandra. Vad \"{a}r sannolikheten f%
\"{o}r att systemet fungerar (svara med tv\aa\ decimaler)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt%
\begin{equation*}
W_{i}=\text{komponent }i\text{ fungerar}
\end{equation*}%
d\aa\ kan funktionssannolikheterna f\"{o}r de tv\aa\ seriesystemen skrivas%
\begin{align*}
P\left( W_{1}\cap W_{2}\cap W_{3}\right) & =%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
=%
%TCIMACRO{\FORMULA{p_{1}\times p_{1}\times p_{1}}{p_{1}^{3}}{evaluate} }%
%BeginExpansion
p_{1}^{3}
%EndExpansion
\\
P\left( W_{4}\cap W_{5}\right) & =%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
=%
%TCIMACRO{\FORMULA{p_{2}\times p_{2}}{p_{2}^{2}}{evaluate}}%
%BeginExpansion
p_{2}^{2}%
%EndExpansion
\end{align*}%
Hela systemet kan nu skrivas som en parallellkoppling av tv\aa\ komponenter
med sannolikheterna $0.729$ och $0.64$ f\"{o}r att fungera. Systemets
funktionssannolikhet blir d\"{a}rf\"{o}r%
\begin{align*}
P\left( \left( W_{1}\cap W_{2}\cap W_{3}\right) \cup \left( W_{4}\cap
W_{5}\right) \right) & =P\left( W_{1}\cap W_{2}\cap W_{3}\right) \\
& +P\left( W_{4}\cap W_{5}\right) \\
& -P\left( \left( W_{1}\cap W_{2}\cap W_{3}\right) \cap \left( W_{4}\cap
W_{5}\right) \right) \\
& =%
%TCIMACRO{\FORMULA{p_{1}\times p_{1}\times p_{1}}{p_{1}^{3}}{evaluate}}%
%BeginExpansion
p_{1}^{3}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{2}\times p_{2}}{p_{2}^{2}}{evaluate}}%
%BeginExpansion
p_{2}^{2}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p_{1}\times p_{1}\times p_{1}}{p_{1}^{3}}{evaluate}}%
%BeginExpansion
p_{1}^{3}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{2}\times p_{2}}{p_{2}^{2}}{evaluate} }%
%BeginExpansion
p_{2}^{2}
%EndExpansion
\\
& =%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{evaluate}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}\left( \func{rand}(10,100)/1000,1\right) $

$p_{2}:=\limfunc{nplaces}\left( \func{rand}(100,200)/1000,1\right) $

$p_{3}:=\limfunc{nplaces}\left( \func{rand}(200,300)/1000,1\right) $

$p_{4}:=\limfunc{nplaces}\left( \func{rand}(300,400)/1000,1\right) $

$\limfunc{svar}:=\limfunc{nplaces}\left( 1-\frac{p_{1}}{1-\left(
1-p_{1}\right) \times \left( 1-p_{2}\right) \times \left( 1-p_{3}\right)
\times \left( 1-p_{4}\right) },2\right) $

\subsubsection{Statement}

I ett system av fyra seriekopplade komponenter 
\begin{equation*}
\begin{tabular}{ccccccccc}
& $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$ &  & $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$ &  & $%
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{evaluate}}%
%BeginExpansion
p_{3}%
%EndExpansion
$ &  & $%
%TCIMACRO{\FORMULA{p_{4}}{p_{4}}{evaluate}}%
%BeginExpansion
p_{4}%
%EndExpansion
$ &  \\ \cline{2-2}\cline{4-4}\cline{6-6}\cline{8-8}
--- & \multicolumn{1}{|c}{1} & \multicolumn{1}{|c}{---} & 
\multicolumn{1}{|c}{2} & \multicolumn{1}{|c}{---} & \multicolumn{1}{|c}{3} & 
\multicolumn{1}{|c}{---} & \multicolumn{1}{|c}{4} & \multicolumn{1}{|c}{---}
\\ \cline{2-2}\cline{4-4}\cline{6-6}\cline{8-8}
\end{tabular}%
\ \ \ 
\end{equation*}%
slutar systemet att fungera n\"{a}r en av komponenterna g\aa r s\"{o}nder
och dessa upph\"{o}r att fungera oberoende av varandra. F\"{o}r
komponenterna g\"{a}ller att $1$ upph\"{o}r att fungera med sannolikheten $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$, komponent $2$ med sannolikheten $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$ \fosv. Om systemet har slutat fungera vad \"{a}r sannolikheten f\"{o}r att
komponent $1$ ej \"{a}r trasig (svara med 2 decimaler).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt 
\begin{align*}
A_{i}& =\text{komponent }i\text{ \"{a}r trasig} \\
B& =\text{systemet fungerar ej}
\end{align*}%
Vi skall ber\"{a}kna 
\begin{align*}
P\left( \complement A_{1}\mid B\right) & =1-P\left( A_{1}\mid B\right) \\
& =1-\frac{P\left( A_{1}\cap B\right) }{P\left( B\right) } \\
\left\{ \text{ty }A_{1}\subseteq B\right\} & =1-\frac{P\left( A_{1}\right) }{%
P\left( B\right) }
\end{align*}%
Sannolikheten i n\"{a}mnaren erh\aa lls, p\aa\ grund av oberoendet, till%
\begin{align*}
P\left( B\right) & =1-P\left( \text{systemet fungerar}\right) \\
& =1-P\left( \complement A_{1}\cap \complement A_{2}\cap \complement
A_{3}\cap \complement A_{4}\right) \\
& =1-P\left( \complement A_{1}\right) P\left( \complement A_{2}\right)
P\left( \complement A_{3}\right) P\left( \complement A_{4}\right) \\
& =1-%
%TCIMACRO{\FORMULA{\left( 1-p_{1}\right) }{1-p_{1}}{evaluate}}%
%BeginExpansion
1-p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{\left( 1-p_{2}\right) }{1-p_{2}}{evaluate}}%
%BeginExpansion
1-p_{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{\left( 1-p_{3}\right) }{1-p_{3}}{evaluate}}%
%BeginExpansion
1-p_{3}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{\left( 1-p_{4}\right) }{1-p_{4}}{evaluate} }%
%BeginExpansion
1-p_{4}
%EndExpansion
\\
& =%
%TCIMACRO{%
%\FORMULA{1-\left( 1-p_{1}\right) \times \left( 1-p_{2}\right) \times \left( 1-p_{3}\right) \times \left( 1-p_{4}\right) }{1-\left( p_{1}-1\right) \left( p_{2}-1\right) \left( p_{3}-1\right) \left( p_{4}-1\right) }{evaluate}}%
%BeginExpansion
1-\left( p_{1}-1\right) \left( p_{2}-1\right) \left( p_{3}-1\right) \left( p_{4}-1\right) %
%EndExpansion
\end{align*}%
H\"{a}rav 
\begin{equation*}
P\left( \complement A_{1}\mid B\right) =%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{evaluate}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\end{equation*}

\section{Question}

\subsection{Comment}

Bayes sats och lagen om total sannolikhet

\subsection{Variant}

\subsubsection{Setup}

$m:=\func{rand}(5,10)$

$p:=\limfunc{nplaces}\left( \func{rand}(400,600)/1000,1\right) $

$\limfunc{svar}:=\limfunc{nplaces}\left( \frac{p}{1\times p+\frac{1}{m}%
\left( 1-p\right) },2\right) $

\subsubsection{Statement}

I ett flervalstest k\"{a}nner en student antingen till det korrekt svaret
eller s\aa\ gissar han. L\aa t $p$ vara sannolikheten att han kan svaret och 
$1-p$ att han m\aa ste gissa. Om studenten m\aa ste gissa bland de $m$
alternativen gissar han r\"{a}tt med sannolikheten $\frac{1}{m}$. Om
studenten gav r\"{a}tt svar vad \"{a}r d\aa\ sannolikheten att studenten
verkligen kunde svaret om $m=%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
$ och $p=%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
$ (svara med 2 decimaler)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

S\"{a}tt 
\begin{align*}
E& =\text{studenten svarar korrekt} \\
F& =\text{studenten kan svaret}
\end{align*}%
d\aa\ erh\aa lls den s\"{o}kta sannolikheten till 
\begin{align*}
P\left( F\mid E\right) & =\frac{P\left( F\cap E\right) }{P\left( E\right) }
\\
& =\frac{P\left( E\mid F\right) P\left( F\right) }{P\left( E\mid F\right)
P\left( F\right) +P\left( E\mid \complement F\right) P\left( \complement
F\right) } \\
& =\frac{1\times 
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
}{1\times 
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
+\frac{1}{%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
}\times \left( 1-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
\right) } \\
& =%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{simplify}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p:=\left( \limfunc{nplaces}\left( \func{rand}(900,999)/1000,2\right) ,%
\limfunc{nplaces}\left( \func{rand}(5,30)/1000,2\right) ,\limfunc{nplaces}%
\left( \func{rand}(2,7)/1000,2\right) \right) $

$\limfunc{svar}:=\limfunc{nplaces}\left( \frac{p_{1}\times p_{3}}{%
p_{1}\times p_{3}+p_{2}\times \left( 1-p_{3}\right) },2\right) $

\subsubsection{Statement}

Ett blodprovstest uppt\"{a}cker med sannolikheten $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$ en sjukdom om den testade personen har sjukdomen. Med sannolikheten $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$ s\"{a}ger testet att personen har sjukdomen n\"{a}r s\aa\ inte \"{a}r
fallet. Andelen individer i befolkningen som har sjukdomen \"{a}r $%
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{evaluate}}%
%BeginExpansion
p_{3}%
%EndExpansion
$. Vad \"{a}r sannolikheten att en person har sjukdomen givet att testet s%
\"{a}ger s\aa\ (svara med tv\aa\ decimaler)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsubsection{Solution}

S\"{a}tt 
\begin{align*}
E& =\text{testresultatet \"{a}r positivt} \\
F& =\text{personen har sjukdomen}
\end{align*}%
d\aa\ erh\aa lls den s\"{o}kta sannolikheten till 
\begin{align*}
P\left( F\mid E\right) & =\frac{P\left( F\cap E\right) }{P\left( E\right) }
\\
& =\frac{P\left( E\mid F\right) P\left( F\right) }{P\left( E\mid F\right)
P\left( F\right) +P\left( E\mid \complement F\right) P\left( \complement
F\right) } \\
& =\frac{%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{simplify}}%
%BeginExpansion
p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{simplify}}%
%BeginExpansion
p_{3}%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{simplify}}%
%BeginExpansion
p_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{simplify}}%
%BeginExpansion
p_{3}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{simplify}}%
%BeginExpansion
p_{2}%
%EndExpansion
\times \left( 1-%
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{simplify}}%
%BeginExpansion
p_{3}%
%EndExpansion
\right) } \\
& =%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{simplify}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{1-p_{1}}{3-p_{1}}$

\subsubsection{Statement}

Du letar efter ett brev fr\aa n skattemyndigheterna och vet att det med lika
sannolikhet skall ligga i en av tre mappar. Antag att du vid en snabb genomg%
\aa ng av en mapp har sannolikheten $p_{i}$ att hitta brevet om det
verkligen ligger i mapp $i$. Om du snabbs\"{o}ker i mapp $1$ och ej finner
brevet d\"{a}r vad \"{a}r sannolikheten att brevet faktiskt finns i mapp $1$
(svara p\aa\ formen $\frac{p}{q}$)?

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\subsubsection{Solution}

S\"{a}tt 
\begin{align*}
F_{i}& =\text{brevet finns i mapp }i\quad i=1,2,3 \\
E& =\text{en snabbs\"{o}kning av mapp }1\text{ ger ej n\aa got brev}
\end{align*}%
d\aa\ erh\aa lls s\"{o}kt sannolikhet till 
\begin{align*}
P\left( F_{1}\mid E\right) & =\frac{P\left( E\mid F_{1}\right) P\left(
F_{1}\right) }{\sum_{i=1}^{3}P\left( E\mid F_{i}\right) P\left( F_{i}\right) 
} \\
& =\frac{\left( 1-p_{1}\right) \frac{1}{3}}{\left( 1-p_{1}\right) \frac{1}{3}%
+1\times \frac{1}{3}+1\times \frac{1}{3}} \\
& =\frac{1-p_{1}}{3-p_{1}}
\end{align*}

\end{document}