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Finish solution for exercise 2

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Andreas Tsouchlos 2025-10-27 10:24:29 +01:00
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src/2025-11-21/presentation.tex
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@ -458,14 +458,40 @@
\item Berechnen Sie die Wahrscheinlichkeit für da Auftreten von
Fehler $C$.
\pause\begin{align*}
P(C) &= P(C\vert AB)P(AB) + P(C\vert A \overline{B})P(A \overline{B})
+ P(C\vert \overline{A}B)P(\overline{A} B)
P(C) &= P(C\vert AB)P(AB) + \cancelto{0}{P(C\vert A \overline{B})}P(A \overline{B})
+ \cancelto{0}{P(C\vert \overline{A}B)}P(\overline{A} B)
+ P(C\vert \overline{A}\overline{B})P(\overline{A}\overline{B}) \\
&= P(C\vert AB)P(AB) + P(C\vert A \overline{B})P(A \overline{B})\\
&= 0.02\cdot 0.01 + 0.01\cdot 0.92 = 0.0094
\end{align*}
\vspace*{-12mm}\pause \item Sie beobachten, dass ein Werkstück den Fehler $C$ hat. Mit
welcher Wahrscheinlichkeit hat es auch Fehler $A$?
\pause\hspace*{-5mm}\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
P(A\vert C) &= \frac{P(AC)}{P(C)}\\[5mm]
P(AC) &= P(ACB) + P(AC \overline{B})\\
&= P(C\vert AB)P(AB) + \cancelto{0}{P(C\vert A \overline{B})}P(A \overline{B})\\
&= 0.02\cdot 0.01 = 0.0002\\[5mm]
P(A\vert C) &= \frac{0.0002}{0.0094} \approx 0.0213
\end{align*}
\end{minipage}%
\hspace*{-10mm}
\begin{minipage}{0.06\textwidth}
\centering
\begin{tikzpicture}
\draw[line width=1pt] (0,0) -- (0,6cm);
\end{tikzpicture}
\end{minipage}%
\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
P(A\vert C) &= \frac{P(C\vert A)P(A)}{P(C)}\\[5mm]
P(C\vert A) &= P(C\vert AB)P(B\vert A)
+ \cancelto{0}{P(C\vert \overline{A} B)}P(\vert \overline{A}B) \\
&= P(C\vert AB)\frac{P(AB)}{P(A)} = 0.02 \cdot \frac{0.01}{0.05} = 0.004\\[5mm]
P(A\vert C) &= \frac{0.004\cdot 0.05}{0.0094} \approx 0.0213
\end{align*}
\end{minipage}
\end{enumerate}
% tex-fmt: on
\end{frame}