Endlicher Wahrscheinlichkeitsraum/Stammbruchsumme/100/Ereignisse/Aufgabe/Lösung

Es ist

1. ${}\mu {\left(\{a,e\}\right)}=f(a)+f(e)={\frac {1}{100}}+{\frac {50}{100}}={\frac {51}{100}}\,,$

2. ${}\mu {\left(\{b,c,e\}\right)}=f(b)+f(c)+f(e)={\frac {4}{100}}+{\frac {20}{100}}+{\frac {50}{100}}={\frac {74}{100}}={\frac {37}{50}}\,,$

3. ${}\mu {\left(\{a,c,d\}\right)}=f(a)+f(c)+f(d)={\frac {1}{100}}+{\frac {20}{100}}+{\frac {25}{100}}={\frac {46}{100}}={\frac {23}{50}}\,,$

4. ${}\mu {\left(\{a,b,d,e\}\right)}=f(a)+f(b)+f(d)+f(e)={\frac {1}{100}}+{\frac {4}{100}}+{\frac {25}{100}}+{\frac {50}{100}}={\frac {80}{100}}={\frac {4}{5}}\,.$

Zur gelösten Aufgabe