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Wiederholung; 456/5
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Scriptim committed Aug 14, 2019
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1 change: 1 addition & 0 deletions Mathe.tex
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\input{Mathe_11_2.tex}
\input{Mathe_12_1.tex}
\input{Mathe_12_2.tex}
\input{Mathe_13_1.tex}

\end{document}
30 changes: 30 additions & 0 deletions Mathe_13_1.tex
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\part{13/1}

\section{Wiederholung}
\begin{itemize}
\item relative Häufigkeit: Aussage über ein schon durchgeführtes Zufallsexperiment
\item Wahrscheinlichkeit: Aussage über zukünftige Zufallsversuche
\end{itemize}
\begin{exercise}{456/5}
\begin{gather*}
F = \{(1, 1), (2, 1), (3, 1), (4, 1)\}
\end{gather*}
\item [a]
\begin{gather*}
E = \{(1, 1), (1, 2), (2, 1)\} \\\\
F \colon \text{Die zweite Kugel trägt eine $1$} \\
E \Leftrightarrow \text{Die Summe der Zahlen ist größer als $3$} \\\\
P(E) = (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{2}{6} \cdot \frac{1}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) = \frac{2}{9} = 0.\overline{2} \\
P(F) = (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) + (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) = P(1) = \frac{2}{6} = \frac{1}{3} = 0.\overline{3}
\end{gather*}
\item [b]
\begin{gather*}
E \cap F = \{(1, 1), (2, 1)\} \\
P(E \cap F) = (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) = \frac{1}{6} = 0.1\overline{6}
\end{gather*}
\item [c]
\begin{gather*}
E \cup F = \{(1, 1), (1, 2), (2, 1), (3, 1), (4, 1)\} \\
P(E \cup F) = (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{2}{6} \cdot \frac{1}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) + (\frac{2}{6} \cdot \frac{2}{6}) + (\frac{1}{6} \cdot \frac{2}{6}) = \frac{7}{18} = 0.3\overline{8}
\end{gather*}
\end{exercise}

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