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Add theory for exercise 1

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Andreas Tsouchlos 2025-10-26 16:27:12 +01:00
parent edbadcff02
commit 1ad4623e22
1 changed files with 148 additions and 5 deletions
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src/2025-11-21/presentation.tex
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@ -43,7 +43,7 @@
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{spy, external, intersections}
\usetikzlibrary{spy, external, intersections, positioning}
%\tikzexternalize[prefix=build/]
\usepackage{pgfplots}
@ -80,6 +80,145 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\frametitle{Bedingte Wahrscheinlichkeiten \& Bayes}
\vspace*{-10mm}
\begin{columns}
\column{\kitthreecolumns}
\begin{itemize}
\item Definition der bedingten Wahrscheinlichkeit
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\item Formel von Bayes
\begin{gather*}
P(A\vert B) = \frac{P(B\vert A) P(A)}{P(B)}
\end{gather*}
\end{itemize}
\column{\kitthreecolumns}
\begin{figure}
\centering
\begin{tikzpicture}
\node[rectangle, minimum width=8cm, minimum height=5cm,
draw, line width=1pt, fill=black!20] at (0,0) {};
\node [circle, minimum size = 4cm,
draw, line width=1pt, fill=KITgreen,
fill opacity = 0.5] at (1.25cm,0) {};
\draw[line width=1pt, fill=KITblue,
fill opacity = 0.5, rounded corners=5mm]
(-2.4cm, -2.25cm) -- (-2.4cm, 2.25cm) -- (1.1cm,0) -- cycle;
\node[left] at (4cm, 2cm) {\Large $\Omega$};
\node at (-1.8cm, 0) {$A$};
\node at (1.8cm, 0) {$B$};
\node at (0, 0) {$AB$};
\end{tikzpicture}
\end{figure}
\end{columns}
\vspace*{1cm}
\pause
\begin{columns}
\column{\kitthreecolumns}
\begin{itemize}
\item Satz der totalen Wahrscheinlichkeit
% tex-fmt: off
\begin{gather*}
\text{Voraussetzungen: }\hspace{5mm} \left\{
\begin{array}{l}
A_1, A_2, \ldots \text{ disjunkt}\\
\displaystyle\sum_{n} A_n = \Omega
\end{array}
\right.\\[1em]
P(B) = \sum_{n} P(B\vert A_n)P(A_n)\\
\end{gather*}
% tex-fmt: on
\end{itemize}
\column{\kitthreecolumns}
\begin{figure}
\centering
\begin{tikzpicture}
\newcommand{\hordist}{1.2cm}
\newcommand{\vertdist}{2cm}
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm] (root) at (0, 0) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and
2.4*\hordist of root] (n1) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and
2.4*\hordist of root] (n2) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and \hordist
of n1] (n11) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and \hordist
of n1] (n12) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and \hordist
of n2] (n21) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and \hordist
of n2] (n22) {};
\draw[-{Latex}, line width=1pt] (root) -- (n1);
\draw[-{Latex}, line width=1pt] (root) -- (n2);
\draw[-{Latex}, line width=1pt] (n1) -- (n11);
\draw[-{Latex}, line width=1pt] (n1) -- (n12);
\draw[-{Latex}, line width=1pt] (n2) -- (n21);
\draw[-{Latex}, line width=1pt] (n2) -- (n22);
\node[left] at ($(root)!0.4!(n1)$) {$P(A_1)$};
\node[right] at ($(root)!0.4!(n2)$) {$P(A_2)$};
\node[left] at ($(n1)!0.4!(n11)$) {$P(B\vert A_1)$};
\node[right] at ($(n1)!0.2!(n12)$) {$P(C\vert A_1)$};
\node[left] at ($(n2)!0.6!(n21)$) {$P(B\vert A_2)$};
\node[right] at ($(n2)!0.4!(n22)$) {$P(C\vert A_2)$};
\node[below] at (n11) {$P(BA_1)$};
\node[below] at (n12) {$P(CA_2)$};
\node[below] at (n21) {$P(BA_1)$};
\node[below] at (n22) {$P(CA_2)$};
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}
\column{\kitthreecolumns}
\begin{greenblock}{Bedingte Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\begin{greenblock}{Formel von Bayes}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\end{greenblock}
\end{columns}
\begin{columns}
\column{\kitonecolumn}
\column{\kitthreecolumns}
\begin{greenblock}{Satz der totalen Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(B) = \sum_{n} P(B\vert A_n)P(A_n)
\end{gather*}
\end{greenblock}
\column{\kitonecolumn}
\end{columns}
\end{frame}
% \begin{frame}{Ereignisse \& Laplace}
% \vspace*{-15mm}
% \begin{itemize}
@ -160,7 +299,8 @@
% \begin{columns}
% \column{\kitthreecolumns}
% \begin{gather*}
% P_r = \frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}
% P_r =
% \frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}
% \end{gather*}
% \column{\kitthreecolumns}
% \begin{lightgrayhighlightbox}
@ -270,7 +410,8 @@
% \begin{align*}
% \mathcal{P}(\Omega) = \{ &\emptyset,
% \mleft\{ A \mright\}, \mleft\{ B \mright\},
% \mleft\{ C \mright\}, \mleft\{ A, B \mright\},\\
% \mleft\{ C \mright\}, \mleft\{ A, B
% \mright\},\\
% &\mleft\{ A, C \mright\},
% \mleft\{ B, C \mright\}, \mleft\{ A, B, C
% \mright\} \}
@ -302,7 +443,8 @@
% \mright) \in \Omega : a_i \neq a_j, i \neq j
% \mright\}\\
% \begin{array}{r}
% \text{Alle Elemente von $\Omega$ unterscheidbar:} \\
% \text{Alle Elemente von $\Omega$
% unterscheidbar:} \\
% \text{Jeweils $L_1, L_2, \ldots, L_M$ der Elemente
% sind gleich:}
% \end{array}
@ -382,7 +524,8 @@
Bei einer Qualitätskontrolle können Werkstücke zwei Fehler
aufweisen: Fehler A, Fehler B, oder
beide Fehler gleichzeitig. Die folgenden Wahrscheinlichkeiten sind bekannt:
beide Fehler gleichzeitig. Die folgenden Wahrscheinlichkeiten
sind bekannt:
\begin{itemize}
\item mit Wahrscheinlichkeit 0,05 hat ein Werkstück den Fehler A
\item mit Wahrscheinlichkeit 0,01 hat ein Werkstück beide Fehler