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10 Commits
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@@ -275,17 +275,17 @@
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\item mindestens ein Ass hat?\pause
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\begin{gather*}
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P(\text{mindestens ein Ass}) = 1 - P(\text{kein Ass})
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= 1 - \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} \approx 0,341
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= 1 - \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} \approx 0{,}341
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\end{gather*}\pause\vspace*{-5mm}
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\item genau ein Ass hat?\pause
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\begin{gather*}
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P(\text{genau ein Ass}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} \approx 0,299
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P(\text{genau ein Ass}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} \approx 0{,}299
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\end{gather*}\pause
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\item mindestens zwei Karten der gleichen Art (“Paar”) hat?\pause
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\begin{align*}
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P(\text{mindestens zwei gleiche Karten}) &= 1 - P(\text{alle Karten unterschiedlich}) \\
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&= 1 - \frac{\text{Anzahl Möglichkeiten mit nur unterschiedlichen Karten}}{\text{Anzahl Möglichkeiten}}\\
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&= 1 - \frac{\binom{13}{5}\cdot 4^5}{\binom{52}{5}} \approx 0,493
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&= 1 - \frac{\binom{13}{5}\cdot 4^5}{\binom{52}{5}} \approx 0{,}493
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\end{align*}
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\end{enumerate}
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% tex-fmt: on
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