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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{Problemdel}

\section{Question}

\subsection{Setup}

Select: 5

\subsection{Comment}

Klassisk sannolikhetsdefinition

\subsection{Variant}

\subsubsection{Setup}

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

\subsubsection{Statement}

Vid tillverkning av en viss typ av flygplansmotorer \"{a}r felsannolikheten $%
p$, dvs\ motorn upph\"{o}r att fungera med denna sannolikhet. Motorer av
denna typ anv\"{a}nds i b\aa de tv\aa -- och fyrmotoriga plan. Ett tv\aa %
motorigt plan m\aa ste ha minst en fungerande motor f\"{o}r att inte st\"{o}%
rta medan ett fyrmotorigt plan kr\"{a}ver minst tv\aa\ fungerande motorer f%
\"{o}r att ej st\"{o}rta (antingen minst en p\aa\ varje vinge eller tv\aa\ p%
\aa\ en vinge). F\"{o}r vilka v\"{a}rden p\aa\ $p$ skulle du helst vilja
befinna dig i det tv\aa motoriga planet. Det kan f\"{o}ruts\"{a}ttas att
motorerna arbetar oberoende av varandra (svara p\aa\ formen $?<p<?$).

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

F\"{o}ljande g\"{a}ller f\"{o}r

\begin{description}
\item[2--motorigt plan] 

\begin{description}
\item d\"{a}r $p_{2}=P\left( \text{kunna flyga}\right) $%
\begin{align*}
p_{2}& =1-P\left( \text{ingen fungerar}\right) \\
& =1-p^{2}
\end{align*}
\end{description}

\item[4--motorigt plan] d\"{a}r $p_{4}=P\left( \text{kunna flyga}\right) $%
\begin{align*}
p_{4}& =1-P\left( \text{mindre \"{a}n 2 fungerar}\right) \\
& =1-P\left( \text{ingen fungerar}\right) -P\left( \text{en fungerar}\right)
\\
& =1-p^{4}-\binom{4}{1}\left( 1-p\right) p^{3}
\end{align*}
\end{description}

Det 2--motoriga planet \"{a}r s\"{a}krare \"{a}n det 4--motoriga om 
\begin{equation*}
p_{2}>p_{4}
\end{equation*}%
vilket ger oss olikheten 
\begin{align*}
1-p^{2}& >1-4\left( 1-p\right) p^{3}-p^{4} \\
1& <4\left( 1-p\right) p+p^{2} \\
3p^{2}-4p+1& <0
\end{align*}%
Om vi ritar upp funktionen $f\left( p\right) =3p^{2}-4p+1$\FRAME{fhF}{%
6.0934cm}{4.3713cm}{0pt}{}{}{gyl_9708.wmf}{\special{language "Scientific
Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file
"F";width 6.0934cm;height 4.3713cm;depth 0pt;original-width
394.125pt;original-height 281.625pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename '../Bilder/Pic-San-Motor.wmf';file-properties
"XNPEU";}} ser vi att denna \"{a}r negativ mellan ett tal $p_{0}$ och $1$. F%
\"{o}r att l\"{o}sa ut den undre gr\"{a}nsen $p_{0}$ har vi att l\"{o}sa
ekvationen 
\begin{equation*}
f\left( p\right) =0
\end{equation*}%
som ger l\"{o}sningarna 
\begin{equation*}
\left\{ \frac{1}{3},1\right\}
\end{equation*}%
Det 2--motoriga planet \"{a}r s\aa ledes s\"{a}krare \"{a}n det 4--motoriga
om felsannolikheten ligger i intervallet $\left( \frac{1}{3},1\right) $. Ett
n\aa got ov\"{a}ntat resultat!

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\frac{1}{\sqrt{\pi n}}$

\subsubsection{Statement}

Ett symmetriskt mynt kastas $2n$ g\aa nger. Uppskatta sannolikheten f\"{o}r
att antalet \textquotedblright gubbe\textquotedblright\ \"{a}r lika med
antalet \textquotedblright krona\textquotedblright\ med hj\"{a}lp av
Stirlings formel $n!=\sqrt{2\pi n}n^{n}e^{-n}$ (svara p\aa\ formen $\frac{p}{%
q}$).

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

S\"{a}tt 
\begin{equation*}
X=\text{totala antalet kast som ger \textquotedblright
gubbe\textquotedblright }
\end{equation*}%
Det finns $\binom{2n}{n}$ gynnsamma sekvenser $\left( X_{1},X_{2},\ldots
,X_{2n}\right) $ dvs sekvenser som inneh\aa ller exakt $n$ stycken gubbe
bland $2^{2n}$ m\"{o}jliga sekvenser. Det f\"{o}rsta talet erh\aa lls ur
formeln f\"{o}r att ta $k$ element ur $m$ d\"{a}r $k=n$ och $m=n$ och det
sista med hj\"{a}lp av multiplikationsprincipen. Sannolikheten kan nu tecknas%
\begin{equation*}
P\left( X=n\right) =\frac{\binom{2n}{n}}{2^{2n}}
\end{equation*}%
och med hj\"{a}lp av Stirlings formel erh\aa lls%
\begin{align*}
& =\frac{\left( 2n\right) !}{n!n!}\frac{1}{2^{2n}} \\
& \approx \frac{\sqrt{2\pi 2n}\left( 2n\right) ^{2n}e^{-2n}}{\sqrt{2\pi n}%
n^{n}e^{-n}\sqrt{2\pi n}n^{n}e^{-n}}\frac{1}{2^{n}} \\
& =\frac{1}{\sqrt{\pi n}}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{01}:=\limfunc{nplaces}(\func{rand}\left( 1,20\right) /1000,3)$

$p_{10}:=\limfunc{nplaces}(\func{rand}\left( 1,20\right) /1000,3)$

$q:=\limfunc{nplaces}(\func{rand}\left( 40,60\right) /100,3)$

$\limfunc{svar}:=(\limfunc{nplaces}\left( \frac{\left( 1-p_{10}\right) q}{%
\left( 1-p_{10}\right) q+p_{01}\left( 1-q\right) },3\right) ,\limfunc{nplaces%
}\left( p_{10}\times q+p_{01}\times \left( 1-q\right) ,3\right) )$

\subsubsection{Statement}

Meddelanden kodade i bin\"{a}ra tecken $0$ och $1$ \"{o}verf\"{o}res i ett
telekommunikationssystem. Signalerna st\"{o}rs av ett brus och d\"{a}rf\"{o}%
r f\"{o}rekommer felaktiga \"{o}verf\"{o}ringar. Ett uts\"{a}nt tecken $0$
mottages som $1$ med sannolikheten $%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
$ och en uts\"{a}nd $1$:a mottages som $0$ med sannolikheten $%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
$. Vidare f\"{o}rekommer tecknen $1$ i proportionen $%
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
$.

\subsubsection{Substatement}

Om $1$ mottagits vad \"{a}r d\aa\ den betingade sannolikheten att $1$ har s%
\"{a}nts?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Substatement}

Hur stor proportion av tecknen \"{o}verf\"{o}rs felaktigt? Med andra ord:\
vad \"{a}r sannolikheten att ett p\aa\ m\aa f\aa\ utvalt tecken \"{a}r
felaktigt mottaget?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

Inf\"{o}r f\"{o}ljande beteckningar: 
\begin{align*}
x_{m}& =\text{den bin\"{a}ra siffran }x\text{ har e\textbf{m}ottagits} \\
x_{s}& =\text{den bin\"{a}ra siffran }x\text{ har \textbf{s}\"{a}nts}
\end{align*}%
Texten ger d\aa\ f\"{o}ljande betingade sannolikheter%
\begin{align*}
P\left( 0_{m}\mid 0_{s}\right) & =%
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate} }%
%BeginExpansion
1-p_{01}
%EndExpansion
\\
P\left( 0_{m}\mid 1_{s}\right) & =%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate} }%
%BeginExpansion
p_{10}
%EndExpansion
\\
P\left( 1_{m}\mid 0_{s}\right) & =%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate} }%
%BeginExpansion
p_{01}
%EndExpansion
\\
P\left( 1_{m}\mid 1_{s}\right) & =%
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate} }%
%BeginExpansion
1-p_{10}
%EndExpansion
\\
P\left( 1_{s}\right) & =%
%TCIMACRO{\FORMULA{q}{q}{evaluate} }%
%BeginExpansion
q
%EndExpansion
\\
P\left( 0_{s}\right) & =%
%TCIMACRO{\FORMULA{1-q}{1-q}{evaluate}}%
%BeginExpansion
1-q%
%EndExpansion
\end{align*}

a) Sannolikheten att en $1$:a har s\"{a}nts n\"{a}r en $1$:a har mottagits
blir d\aa\ 
\begin{align*}
P\left( 1_{s}\mid 1_{m}\right) & =\frac{P\left( 1_{s}\cap 1_{m}\right) }{%
P\left( 1_{m}\right) }=\frac{P\left( 1_{m}\mid 1_{s}\right) P\left(
1_{s}\right) }{P\left( 1_{m}\mid 1_{s}\right) P\left( 1_{s}\right) +P\left(
1_{m}\mid 0_{s}\right) P\left( 0_{s}\right) } \\
& =\frac{%
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-q}{1-q}{evaluate}}%
%BeginExpansion
1-q%
%EndExpansion
} \\
& =%
%TCIMACRO{%
%\FORMULA{\frac{\left( 1-p_{10}\right) q}{\left( 1-p_{10}\right) q+p_{01}\left( 1-q\right) }}{q\frac{p_{10}-1}{q\left( p_{10}-1\right) +p_{01}\left( q-1\right) }}{evaluate}}%
%BeginExpansion
q\frac{p_{10}-1}{q\left( p_{10}-1\right) +p_{01}\left( q-1\right) }%
%EndExpansion
\end{align*}

b) Det finns tv\aa\ m\"{o}jligheter att g\"{o}ra fel\"{o}verf\"{o}ringar 
\begin{align*}
P\left( \text{fel\"{o}verf\"{o}ring}\right) & =P\left( 1_{s}\cap
0_{m}\right) +P\left( 0_{s}\cap 1_{m}\right) =P\left( 0_{m}\mid 1_{s}\right)
P\left( 1_{s}\right) +P\left( 1_{m}\mid 0_{s}\right) P\left( 0_{s}\right) \\
& =%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-q}{1-q}{evaluate}}%
%BeginExpansion
1-q%
%EndExpansion
=%
%TCIMACRO{%
%\FORMULA{p_{10}\times q+p_{01}\times \left( 1-q\right) }{qp_{10}-p_{01}\left( q-1\right) }{evaluate}}%
%BeginExpansion
qp_{10}-p_{01}\left( q-1\right) %
%EndExpansion
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

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

\subsubsection{Statement}

Till en reparationsverkstad inkommer $3$ mobiltelefoner och deras
telefonkort tas ut f\"{o}re reparation. P\aa\ grund av ett misstag kommer
kortens identifikationsnummer bort. Efter avslutad reparation paras d\"{a}rf%
\"{o}r kort och telefon godtyckligt ihop. Vad \"{a}r sannolikheten att ingen
av \"{a}garna kan anv\"{a}nda sin telefon (till varje kort h\"{o}r en
entydig pinkod som bara \"{a}garen k\"{a}nner till) (svara p\aa\ formen $%
\frac{p}{q}$)?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

S\"{a}tt 
\begin{equation*}
E_{i}=\text{Telefon och kort passar f\"{o}r telefon }i\text{\quad }i=1,2,3
\end{equation*}%
Vi s\"{o}ker nu 
\begin{align*}
P\left( \complement E_{1}\cap \complement E_{2}\cap \complement E_{3}\right)
& =P\left( \text{inget kort paras med r\"{a}tt telefon}\right) \\
& =1-P\left( E_{1}\cup E_{2}\cup E_{3}\right)
\end{align*}%
Nu g\"{a}ller att (visa det!) 
\begin{align*}
P\left( E_{1}\cup E_{2}\cup E_{3}\right) & =P\left( E_{1}\right) +P\left(
E_{2}\right) +P\left( E_{3}\right) \\
& -P\left( E_{1}\cap E_{2}\right) -P\left( E_{1}\cap E_{3}\right) -P\left(
E_{2}\cap E_{3}\right) \\
& +P\left( E_{1}\cap E_{2}\cap E_{3}\right)
\end{align*}%
och via har d\"{a}rf\"{o}r att best\"{a}mma f\"{o}ljande tre typer av
sannolikheter varav den f\"{o}rsta \"{a}r trivial 
\begin{equation*}
P\left( E_{i}\right) =\frac{1}{3}
\end{equation*}%
f\"{o}r den andra finner vi 
\begin{align*}
P\left( E_{1}\cap E_{2}\right) & =P\left( E_{2}\mid E_{1}\right) P\left(
E_{1}\right) =\frac{1}{2}\frac{1}{3} \\
& =\frac{1}{6}
\end{align*}%
och de tv\aa\ andra paren ger samma resultat. Den tredje och sista
sannolikheten erh\aa lls genom upprepad betingning till 
\begin{align*}
P\left( E_{1}\cap E_{2}\cap E_{3}\right) & =P\left( E_{3}\mid E_{1}\cap
E_{2}\right) P\left( E_{2}\mid E_{1}\right) P\left( E_{1}\right) =\frac{1}{1}%
\frac{1}{2}\frac{1}{3} \\
& =\frac{1}{6}\text{.}
\end{align*}%
Den s\"{o}kta sannolikheten blir nu 
\begin{align*}
P\left( \complement E_{1}\cap \complement E_{2}\cap \complement E_{3}\right)
& =1-P\left( E_{1}\cup E_{2}\cup E_{3}\right) \\
& =1-\frac{1}{3}-\frac{1}{3}-\frac{1}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}-%
\frac{1}{6} \\
& =\frac{1}{3}\text{.}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p:=\limfunc{nplaces}(\func{rand}\left( 5,8\right) /10,1)$

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

\subsubsection{Statement}

Tennisspelarna $A$ och $B$ skall spela en match mot varandra. Man kan anta
att sannolikheten att $A$ vinner ett set \"{a}r $%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
$ oberoende av utg\aa ngen av tidigare set. Matchen slutar n\"{a}r en av
spelarna vunnit tv\aa\ set.

\subsubsection{Substatement}

Ber\"{a}kna sannolikheten f\"{o}r att matchen slutar efter $2$ set (svara
med 2 decimaler).

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Substatement}

Ber\"{a}kna sannolikheten f\"{o}r att matchen slutar efter $3$ set (svara
med 2 decimaler).

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Substatement}

Ber\"{a}kna sannolikheten f\"{o}r att $A$ vinner (svara med 2 decimaler).

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

S\"{a}tt 
\begin{eqnarray*}
A &=&A\text{ vinner ett set} \\
B &=&B\text{ vinner ett set} \\
C_{i} &=&\text{matchen slutar efter }i\text{ set\quad }i=2,3
\end{eqnarray*}%
d\aa\ g\"{a}ller f\"{o}r tv\aa\ set 
\begin{align*}
P\left( C_{2}\right) & =P\left( \left\{ AA,BB\right\} \right) =P\left(
AA\right) +P\left( BB\right) \\
& =P\left( A\right) ^{2}+P\left( B\right) ^{2} \\
& =%
%TCIMACRO{\FORMULA{p^{2}}{p^{2}}{evaluate}}%
%BeginExpansion
p^{2}%
%EndExpansion
+%
%TCIMACRO{%
%\FORMULA{\left( 1-p\right) ^{2}}{\left( p-1\right) ^{2}}{evaluate} }%
%BeginExpansion
\left( p-1\right) ^{2}
%EndExpansion
\\
& =%
%TCIMACRO{%
%\FORMULA{p^{2}+\left( 1-p\right) ^{2}}{\left( p-1\right) ^{2}+p^{2}}{evaluate}}%
%BeginExpansion
\left( p-1\right) ^{2}+p^{2}%
%EndExpansion
\end{align*}%
och f\"{o}r tre set%
\begin{align*}
P\left( C_{3}\right) & =P\left( \left\{ ABA,BAA,BAB,ABB\right\} \right)
=2P\left( A\right) ^{2}P\left( B\right) +2P\left( A\right) P\left( B\right)
^{2} \\
& =2\times 
%TCIMACRO{\FORMULA{p^{2}}{p^{2}}{evaluate}}%
%BeginExpansion
p^{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p}{1-p}{evaluate}}%
%BeginExpansion
1-p%
%EndExpansion
+2\times 
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
\times 
%TCIMACRO{%
%\FORMULA{\left( 1-p\right) ^{2}}{\left( p-1\right) ^{2}}{evaluate} }%
%BeginExpansion
\left( p-1\right) ^{2}
%EndExpansion
\\
& =%
%TCIMACRO{%
%\FORMULA{2\times p^{2}\times \left( 1-p\right) +2\times p\times \left( 1-p\right) ^{2}}{2p\left( p-1\right) ^{2}-2p^{2}\left( p-1\right) }{evaluate}}%
%BeginExpansion
2p\left( p-1\right) ^{2}-2p^{2}\left( p-1\right) %
%EndExpansion
\text{.}
\end{align*}

Den senare f\"{o}ljer av att endast sekvenserna $ABA$, $BAA$ ger att $A$
vinner och motsvarande f\"{o}r $B$ \"{a}r $BAB$, $ABB$.

Till sist har vi att%
\begin{eqnarray*}
P\left( A\text{ vinner}\right) &=&P\left( \left\{ AA,BAA,ABA\right\} \right)
\\
&=&P\left( AA\right) +P\left( BAA\right) +P\left( ABA\right) \\
&=&%
%TCIMACRO{\FORMULA{p^{2}}{p^{2}}{evaluate}}%
%BeginExpansion
p^{2}%
%EndExpansion
+%
%TCIMACRO{%
%\FORMULA{2\times \left( 1-p\right) \times p^{2}}{-2p^{2}\left( p-1\right) }{evaluate} }%
%BeginExpansion
-2p^{2}\left( p-1\right) 
%EndExpansion
\\
&=&%
%TCIMACRO{%
%\FORMULA{p^{2}+2\times \left( 1-p\right) \times p^{2}}{p^{2}-2p^{2}\left( p-1\right) }{evaluate}}%
%BeginExpansion
p^{2}-2p^{2}\left( p-1\right) %
%EndExpansion
\end{eqnarray*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\ln \left( 1-x\right) <-x$

\subsubsection{Statement}

En v\"{a}gtunnel passeras under en timme av $n_{1}$ bilar som passerar
tunneln p\aa\ $t_{1}$ minuter, $n_{2}$ bilar som passerar tunneln p\aa\ $%
t_{2}$ minuter osv\ till att $n_{k}$ bilar passerar tunneln p\aa\ $t_{k}$
minuter. H\"{a}ndelserna att de olika bilarna passerar tunneln \"{a}r
oberoende. Om $p$ \"{a}r sannolikheten att tunneln i ett visst \"{o}gonblick 
\"{a}r tom visa att 
\begin{equation*}
p<e^{-\frac{1}{60}\sum_{i=1}^{k}n_{i}t_{i}}\text{.}
\end{equation*}%
Vilken olikhet l\"{o}ser problemet?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

S\"{a}tt 
\begin{equation*}
A_{ij}=\text{bil med passeringstiden }t_{j}\text{ \"{a}r i tunneln }%
i=1,\ldots ,n_{j},j=1,\ldots ,k
\end{equation*}%
D\aa\ g\"{a}ller att sannolikheten f\"{o}r ingen bil i tunneln kan skrivas 
\begin{align*}
p& =P\left( \bigcap_{j=1}^{k}\bigcap_{i=1}^{n_{j}}\complement A_{ij}\right)
\\
& =\prod_{j=1}^{k}\prod_{i=1}^{n_{j}}\left( 1-P\left( A_{ij}\right) \right)
\\
& =\prod_{j=1}^{k}\prod_{i=1}^{n_{j}}\left( 1-\frac{t_{j}}{60}\right) \\
& =e^{\sum_{j=1}^{k}\sum_{i=1}^{n_{j}}\ln \left( 1-\frac{t_{j}}{60}\right) }
\end{align*}%
Nu g\"{a}ller olikheten 
\begin{equation*}
\ln \left( 1-x\right) <-x
\end{equation*}%
varf\"{o}r 
\begin{align*}
p& <e^{-\sum_{j=1}^{k}\sum_{i=1}^{n_{j}}\frac{t_{j}}{60}} \\
& =e^{-\frac{1}{60}\sum_{j=1}^{k}n_{j}t_{j}}\text{.}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{01}:=\limfunc{nplaces}(\func{rand}\left( 15,25\right) /100,2)$

$p_{10}:=\limfunc{nplaces}(\func{rand}\left( 5,15\right) /100,2)$

$q_{1}:=\limfunc{nplaces}(\func{rand}\left( 25,35\right) /100,2)$

$q_{2}:=\limfunc{nplaces}(\func{rand}\left( 25,35\right) /100,2)$

$q_{3}:=\limfunc{nplaces}(1-q_{1}-q_{2},1)$

$l:=\left( 1-p_{01}\right) ^{2}p_{01}\left( 1-p_{10}\right)
p_{10}q_{1}+p_{10}^{2}\left( 1-p_{10}\right) p_{01}\left( 1-p_{01}\right)
q_{2}+p_{10}^{2}\left( 1-p_{10}\right) ^{2}\left( 1-p_{01}\right) q_{3}$

$s_{1}:=\limfunc{nplaces}\left( \frac{\left( 1-p_{01}\right)
^{2}p_{01}\left( 1-p_{10}\right) p_{10}q_{1}}{l},3\right) $

$s_{2}:=\limfunc{nplaces}\left( \frac{p_{10}^{2}\left( 1-p_{10}\right)
p_{01}\left( 1-p_{01}\right) q_{2}}{l},3\right) $

$s_{3}:=\limfunc{nplaces}\left( \frac{p_{10}^{2}\left( 1-p_{10}\right)
^{2}\left( 1-p_{01}\right) q_{3}}{l},3\right) $

$\limfunc{svar}:=\left( s_{1},s_{2},s_{3},00110\right) $

\subsubsection{Statement}

I ett kommunikationssystem \"{o}verf\"{o}rs bin\"{a}ra tecken, $0$ och $1$,
med sannolikheten $%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
$ respektive $%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
$ f\"{o}r felaktig \"{o}verf\"{o}ring. Antag att sekvenserna $00110$, $11010$
och $11101$ s\"{a}nds med sannolikheterna $%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate}}%
%BeginExpansion
q_{1}%
%EndExpansion
$, $%
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate}}%
%BeginExpansion
q_{2}%
%EndExpansion
$ respektive $%
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate}}%
%BeginExpansion
q_{3}%
%EndExpansion
$. Om sekvensen $01100$ mottages vilken av de m\"{o}jliga sekvenserna f\aa r
st\"{o}rst betingad sannolikhet?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

S\"{a}tt 
\begin{align*}
A& =01100\text{ har mottagits} \\
B& =00110\text{ har s\"{a}nts} \\
C& =11010\text{ har s\"{a}nts} \\
D& =11101\text{ har s\"{a}nts}
\end{align*}%
d\aa\ g\"{a}ller att lagen om total sannolikhet ger 
\begin{align*}
P\left( A\right) & =P\left( A\mid B\right) P\left( B\right) +P\left( A\mid
C\right) P\left( C\right) +P\left( A\mid D\right) P\left( D\right) \\
& =%
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate} }%
%BeginExpansion
q_{1}
%EndExpansion
\\
& +%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate} }%
%BeginExpansion
q_{2}
%EndExpansion
\\
& +%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate} }%
%BeginExpansion
q_{3}
%EndExpansion
\\
& =%
%TCIMACRO{\FORMULA{l}{l}{evaluate}}%
%BeginExpansion
l%
%EndExpansion
\end{align*}

D\"{a}refter erh\aa lls med hj\"{a}lp av Bayes sats att 
\begin{align*}
P\left( B\mid A\right) & =\frac{P\left( B\cap A\right) }{P\left( A\right) }=%
\frac{P\left( A\mid B\right) P\left( B\right) }{P\left( A\right) }=\frac{%
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate}}%
%BeginExpansion
q_{1}%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{l}{l}{evaluate}}%
%BeginExpansion
l%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{s_{1}}{s_{1}}{evaluate} }%
%BeginExpansion
s_{1}
%EndExpansion
\\
P\left( C\mid A\right) & =\frac{P\left( C\cap A\right) }{P\left( A\right) }=%
\frac{P\left( A\mid C\right) P\left( C\right) }{P\left( A\right) }=\frac{%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate}}%
%BeginExpansion
q_{2}%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{l}{l}{evaluate}}%
%BeginExpansion
l%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{s_{2}}{s_{2}}{evaluate} }%
%BeginExpansion
s_{2}
%EndExpansion
\\
P\left( D\mid A\right) & =\frac{P\left( D\cap A\right) }{P\left( A\right) }=%
\frac{P\left( A\mid D\right) P\left( D\right) }{P\left( A\right) }=\frac{%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{10}}{1-p_{10}}{evaluate}}%
%BeginExpansion
1-p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{01}}{1-p_{01}}{evaluate}}%
%BeginExpansion
1-p_{01}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate}}%
%BeginExpansion
q_{3}%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{l}{l}{evaluate}}%
%BeginExpansion
l%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{s_{3}}{s_{3}}{evaluate}}%
%BeginExpansion
s_{3}%
%EndExpansion
\end{align*}%
Sekvensen $00110$ har d\"{a}rf\"{o}r st\"{o}rst betingad sannolikhet.

\subsection{Variant}

\subsubsection{Setup}

$p_{00}:=\limfunc{nplaces}(\func{rand}\left( 40,50\right) /100,2)$

$p_{01}:=\limfunc{nplaces}(\func{rand}\left( 45,55\right) /100,2)$

$p_{02}:=\limfunc{nplaces}(1-p_{00}-p_{01},2)$

$p_{10}:=\limfunc{nplaces}(\func{rand}\left( 3,7\right) /100,2)$

$p_{11}:=\limfunc{nplaces}(\func{rand}\left( 65,65\right) /100,2)$

$p_{12}:=\limfunc{nplaces}(1-p_{10}-p_{11},2)$

$p_{20}:=\limfunc{nplaces}(\func{rand}\left( 0,3\right) /100,2)$

$p_{21}:=\limfunc{nplaces}(\func{rand}\left( 45,55\right) /100,2)$

$p_{22}:=\limfunc{nplaces}(1-p_{20}-p_{21},2)$

$q_{1}:=\limfunc{nplaces}(\func{rand}\left( 7,13\right) /100,2)$

$q_{2}:=\limfunc{nplaces}(\func{rand}\left( 35,45\right) /100,2)$

$q_{3}:=\limfunc{nplaces}(1-q_{1}-q_{2},2)$

$s_{1}:=\limfunc{nplaces}\left( p_{22}\times q_{3}+p_{12}\times
q_{2}+p_{02}\times q_{1},2\right) $

$s_{2}:=\limfunc{nplaces}\left( \frac{p_{22}\times q_{3}}{p_{22}\times
q_{3}+p_{12}\times q_{2}+p_{02}\times q_{1}},2\right) $

$\limfunc{svar}:=\left( s_{1},s_{2}\right) $

\subsubsection{Statement}

Individerna i ett samh\"{a}lle delades in i tre klasser -- h\"{a}r kallade
A, E och K -- och d\"{a}refter noterade man klasstillh\"{o}righeten f\"{o}r m%
\"{a}n och deras vuxna s\"{o}ner varvid man erh\"{o}ll f\"{o}ljande tabell 
\begin{equation*}
\begin{tabular}{ccccc}
&  & \multicolumn{3}{c}{\textbf{Son}} \\ 
&  & $K_{2}$ & $E_{2}$ & $A_{2}$ \\ 
& $K_{1}$ & $%
%TCIMACRO{\FORMULA{p_{00}}{p_{00}}{evaluate}}%
%BeginExpansion
p_{00}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{01}}{p_{01}}{evaluate}}%
%BeginExpansion
p_{01}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{02}}{p_{02}}{evaluate}}%
%BeginExpansion
p_{02}%
%EndExpansion
$ \\ 
\textbf{Far} & $E_{1}$ & $%
%TCIMACRO{\FORMULA{p_{10}}{p_{10}}{evaluate}}%
%BeginExpansion
p_{10}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{11}}{p_{11}}{evaluate}}%
%BeginExpansion
p_{11}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate}}%
%BeginExpansion
p_{12}%
%EndExpansion
$ \\ 
& $A_{1}$ & $%
%TCIMACRO{\FORMULA{p_{20}}{p_{20}}{evaluate}}%
%BeginExpansion
p_{20}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{21}}{p_{21}}{evaluate}}%
%BeginExpansion
p_{21}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{p_{22}}{p_{22}}{evaluate}}%
%BeginExpansion
p_{22}%
%EndExpansion
$%
\end{tabular}%
\ \ 
\end{equation*}%
Tabellen skall l\"{a}sas 
\begin{equation*}
P\left( A_{2}\mid E_{1}\right) =P\left( \text{sonen tillh\"{o}r klass }A\mid 
\text{fadern tillh\"{o}r klass }E\right) =%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate}}%
%BeginExpansion
p_{12}%
%EndExpansion
\text{.}
\end{equation*}%
(Unders\"{o}kningen utf\"{o}rdes i England och Wales av Glass och Hall 1954 f%
\"{o}r att utr\"{o}na i hur stor utstr\"{a}ckning byte av klasstillh\"{o}%
righet f\"{o}rekom. Man ville f\aa\ svar p\aa\ fr\aa gor av typen: Vilken
chans har en arbetare att arbeta sig upp?) I unders\"{o}kningen f\"{o}%
rdelade sig f\"{a}derna enligt 
\begin{align*}
P\left( \text{Klass }K\right) & =%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate} }%
%BeginExpansion
q_{1}
%EndExpansion
\\
P\left( \text{Klass }E\right) & =%
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate} }%
%BeginExpansion
q_{2}
%EndExpansion
\\
P\left( \text{Klass }A\right) & =%
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate}}%
%BeginExpansion
q_{3}%
%EndExpansion
\end{align*}%
(svara med tv\aa\ decimaler).

\subsubsection{Substatement}

Vad \"{a}r sannolikheten att en son tillh\"{o}r klass $A$?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Substatement}

Om en son tillh\"{o}r klass $A$ vad \"{a}r sannolikheten att \"{a}ven fadern
tillh\"{o}r klass $A$?

\paragraph{Choices}

InputField(MATH)

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

\subsubsection{Solution}

Sannolikheten att en son tillh\"{o}r klass $A$ blir 
\begin{align*}
P\left( A_{2}\right) & =P\left( A_{2}\cap A_{1}\right) +P\left( A_{2}\cap
E_{1}\right) +P\left( A_{2}\cap K_{1}\right) \\
& =P\left( A_{2}\mid A_{1}\right) P\left( A_{1}\right) +P\left( A_{2}\mid
E_{1}\right) P\left( E_{1}\right) +P\left( A_{2}\mid K_{1}\right) P\left(
K_{1}\right) \\
& =%
%TCIMACRO{\FORMULA{p_{22}}{p_{22}}{evaluate}}%
%BeginExpansion
p_{22}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate}}%
%BeginExpansion
q_{3}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{12}}{p_{12}}{evaluate}}%
%BeginExpansion
p_{12}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate}}%
%BeginExpansion
q_{2}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{p_{02}}{p_{02}}{evaluate}}%
%BeginExpansion
p_{02}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate} }%
%BeginExpansion
q_{1}
%EndExpansion
\\
& =%
%TCIMACRO{\FORMULA{s_{1}}{s_{1}}{evaluate}}%
%BeginExpansion
s_{1}%
%EndExpansion
\end{align*}%
Sannolikheten att \"{a}ven fadern tillh\"{o}r klass $A$ blir 
\begin{align*}
P\left( A_{1}\mid A_{2}\right) & =\frac{P\left( A_{2}\mid A_{1}\right)
P\left( A_{1}\right) }{P\left( A_{2}\right) } \\
& =\frac{%
%TCIMACRO{\FORMULA{p_{22}}{p_{22}}{evaluate}}%
%BeginExpansion
p_{22}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{q_{3}}{q_{3}}{evaluate}}%
%BeginExpansion
q_{3}%
%EndExpansion
}{%
%TCIMACRO{%
%\FORMULA{p_{22}\times q_{3}+p_{12}\times q_{2}+p_{02}\times q_{1}}{q_{1}p_{02}+q_{2}p_{12}+q_{3}p_{22}}{evaluate}}%
%BeginExpansion
q_{1}p_{02}+q_{2}p_{12}+q_{3}p_{22}%
%EndExpansion
} \\
& =%
%TCIMACRO{\FORMULA{s_{2}}{s_{2}}{evaluate}}%
%BeginExpansion
s_{2}%
%EndExpansion
\end{align*}

\end{document}