Wiki-Quellcode von Lösung Fruchgummis

Version 15.1 von Stefan Martin am 2025/12/17 13:34

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Stefan Martin 12.1 1 Die Wahrscheinlichkeit für das Ereignis E = "kein gelbes Fruchtgummi" lässt sich z. B. berechnen, in dem die Wahrscheinlichkeit des sog. Gegenereignisses ermittelt und diese schlussendlich von 1 subtrahiert wird:
Stefan Martin 11.1 2
Stefan Martin 1.1 3 {{formula}}
Stefan Martin 8.1 4 \begin{aligned}
Stefan Martin 10.1 5 P(E) & = 1-(P(\text{"gelb, gelb"}) + P(\text{"gelb, rot"}) + P(\text{"gelb, weiß"}) + P(\text{"gelb, grün"}) ) \\
Stefan Martin 9.1 6 & = 1- (313⋅212+2⋅313⋅712+2⋅313⋅212+2⋅313⋅112) \\
7 & = 1526 \\
Stefan Martin 8.1 8 \end{aligned}
Stefan Martin 1.1 9 {{/formula}}
Stefan Martin 11.1 10
Stefan Martin 1.1 11 Die Aussage ist also richtig.
Stefan Martin 13.1 12
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21 \node {}
22 child { node {Rot}
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24 child { node {Rot} edge from parent node[left] {$\frac{6}{16}$} }
25 child { node {Grün} edge from parent node[right] {$\frac{1}{16}$} }
26 child { node {Gelb} edge from parent node[right] {$\frac{3}{16}$} }
27 child { node {Weiß} edge from parent node[right] {$\frac{2}{16}$} }
28 }
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31 child { node {Rot} edge from parent node[left] {$\frac{7}{16}$} }
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34 }
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39 child { node {Weiß} edge from parent node[right] {$\frac{2}{16}$} }
40 }
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43 child { node {Rot} edge from parent node[left] {$\frac{7}{16}$} }
44 child { node {Grün} edge from parent node[right] {$\frac{1}{16}$} }
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46 };
47
48 \end{tikzpicture}
49
50