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Baumdiagramme - Wahrscheinlichkeiten; 455/3; 455/4
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Scriptim committed Jun 20, 2019
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Expand Up @@ -1241,3 +1241,44 @@ \section{Mehrstufige Zufallsversuche - Baumdiagramme}
\text{o. Z.} \quad P((r, r), (r, b)) = \frac{6}{7}
\end{gather*}
\end{exercise}
\subsection{Baumdiagramme - Wahrscheinlichkeiten}
\begin{gather*}
\text{(Würfeln)} \\\\
p(1) = p(2) = \frac{1}{6} \qquad \text{Einzel-/Zweigwahrscheinlichkeit} \\
P(1, 1, 2) = P((1~1~2), (1~2~1), (2~1~1)) = 3 \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{72}
\end{gather*}
\begin{exercise}{455/3}
\begin{gather*}
p(0) = \frac{1}{4} \qquad p(1) = \frac{3}{4}
\end{gather*}
\item [a]
\begin{gather*}
P(0, 0, 0) = (\frac{1}{4})^3 = \frac{1}{64} \approx 0.02 \\
\Leftrightarrow \text{ mindestens ein Patient wird geheilt}
\end{gather*}
\item [b]
\begin{gather*}
P(1, 0, 0) = P((1~0~0), (0~1~0), (0~0~1)) = 3 \cdot (\frac{1}{4})^2 \cdot \frac{3}{4} = \frac{9}{64} \approx 0.14 \\
\Leftrightarrow \text{ kein Patient oder mehr als ein Patient werden geheilt}
\end{gather*}
\item [c]
\begin{gather*}
P(1, 1, 0) = P((1~1~0), (1~0~1), (0~1~1)) = 3 \cdot \frac{1}{4} \cdot (\frac{3}{4})^2 = \frac{27}{64} \approx 0.42 \\
\Leftrightarrow \text{ alle Patienten oder weniger als zwei Patienten werden geheilt}
\end{gather*}
\item [d]
\begin{gather*}
P((0, 0, 0), (1, 0, 0), (1, 1, 0)) = \frac{1}{64} + \frac{9}{64} + \frac{27}{64} = \frac{37}{64} \approx 0.58 \\
\Leftrightarrow \text{ alle Patienten werden geheilt}
\end{gather*}
\end{exercise}
\begin{exercise}{455/4}
\begin{gather*}
P(1) = P((1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)) = \frac{11}{36} \\
P(2) = P((2, 2), (2, 3), (2, 4), (2, 5), (2, 6)) = \frac{9}{36} = \frac{1}{4} \\
P(3) = P((3, 3), (3, 4), (3, 5), (3, 6)) = \frac{7}{36} \\
P(4) = P((4, 4), (4, 5), (4, 6)) = \frac{5}{36} \\
P(5) = P((5, 5), (5, 6)) = \frac{3}{36} = \frac{1}{12} \\
P(6) = P((6, 6)) = \frac{1}{36} \\
\end{gather*}
\end{exercise}

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