Wiki-Quellcode von Lösung Stochastische Unabhängigkeit Vierfeldertafel
Version 1.7 von thomashermann am 2026/05/12 14:51
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| author | version | line-number | content |
|---|---|---|---|
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1.1 | 1 | Für die Ereignisse {{formula}}M=\{ist\,männlich\}{{/formula}} und {{formula}}KI=\{benutzt\,Künstliche\,Intelligenz\}{{/formula}} gilt: |
| 2 | (%class=abc%) | ||
| |
1.2 | 3 | 1. die Ereignisse M und KI sind stochastisch unabhängig |
| |
1.1 | 4 | (%class="border slim"%) |
| 5 | ||={{formula}}M{{/formula}}|={{formula}}\overline{M}{{/formula}}| | ||
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1.2 | 6 | |={{formula}}KI{{/formula}}|{{formula}}0,25{{/formula}}|{{formula}}0,11{{/formula}}|{{formula}}0,36{{/formula}} |
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1.1 | 7 | |={{formula}}\overline{KI}{{/formula}}|{{formula}}0,45{{/formula}}|{{formula}}0,19{{/formula}}|{{formula}}0,64{{/formula}} |
| 8 | | |{{formula}}0,7{{/formula}}|{{formula}}0,3{{/formula}}|{{formula}}1{{/formula}} | ||
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1.6 | 9 | |
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1.5 | 10 | Es gilt; |
| 11 | {{formula}}P(M \cap \overline{KI})=0,7-0,25=0,45{{/formula}} | ||
| 12 | {{formula}}P(\overline{M})=1-0,7=0,3{{/formula}} | ||
| 13 | {{formula}}P(M)\cdot P(KI)=P(M \cap KI){{/formula}} | ||
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1.7 | 14 | {{formula}}\Rightarrow P(KI)=\frac{P(M \cap KI)}{P(M)}{{/formula}} |
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1.5 | 15 | |
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1.2 | 16 | 1. die Ereignisse M und KI sind stochastisch abhängig. |
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1.1 | 17 | (%class="border slim"%) |
| 18 | ||={{formula}}M{{/formula}}|={{formula}}\overline{M}{{/formula}}| | ||
| 19 | |={{formula}}KI{{/formula}}|{{formula}}0,25{{/formula}}|| | ||
| 20 | |={{formula}}\overline{KI}{{/formula}}||| | ||
| 21 | | |{{formula}}0,7{{/formula}}||1 | ||
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1.3 | 22 |