V osudí 2

V osudí je 5 bílých a 9 černých. Namátkou vybereme tři koule. Jaká je pravděpodobnost, že
a) vybrané koule nebudou stejné barvy,
b) mezi nimi budou aspoň dvě černé?

Správná odpověď:

p₁ =  0,7418
p₂ =  0,7253

Postup správného řešení:

$$\begin{gathered} C_3(14)=\left(\genfrac{}{}{0ex}{}{14}{3}\right)=\frac{14!}{3!(14-3)!}=\frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}=364 \\ {C}_2(5)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5!}{2!(5-2)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {C}_1(9)=\left(\genfrac{}{}{0ex}{}{9}{1}\right)=\frac{9!}{1!(9-1)!}=\frac{9}{1}=9 \\ {C}_1(5)=\left(\genfrac{}{}{0ex}{}{5}{1}\right)=\frac{5!}{1!(5-1)!}=\frac{5}{1}=5 \\ {C}_2(9)=\left(\genfrac{}{}{0ex}{}{9}{2}\right)=\frac{9!}{2!(9-2)!}=\frac{9\cdot 8}{2\cdot 1}=36 \\ {p}_1=\frac{\left(\genfrac{}{}{0ex}{}{5}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{1}\right)+\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{2}\right)}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{10\cdot 9+5\cdot 36}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{135}{182}=0,7418 \end{gathered}$$

$$\begin{gathered} C_0(5)=\left(\genfrac{}{}{0ex}{}{5}{0}\right)=\frac{5!}{0!(5-0)!}=\frac{1}{1}=1 \\ {C}_3(9)=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=\frac{9!}{3!(9-3)!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=84 \\ {p}_2=\frac{\left(\genfrac{}{}{0ex}{}{5}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{3}\right)+\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{2}\right)}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{1\cdot 84+5\cdot 36}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{66}{91}=0,7253 \end{gathered}$$

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