Pravděpodobnost soutěž

Ve třídě je 10 žáků, z toho 8 dívek a dva kluci. Chceme vybrat tři na soutěž. Jaká je pravděpodobnost, že to budou:
a) 2 dívky a 1 kluk
b) 1 holka a 2 kluci
c) 3 dívky
d) 3 kluci
e) alespoň 2 dívky

Správná odpověď:

a =  0,4667
b =  0,0667
c =  0,4667
d =  0
e =  0,9333

Postup správného řešení:

$$\begin{gathered} C_2(8)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{2!(8-2)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {C}_1(2)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=\frac{2!}{1!(2-1)!}=\frac{2}{1}=2 \\ {C}_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ {n}_1=8 \\ {n}_2=2 \\ a=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{28\cdot 2}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0,4667 \end{gathered}$$

$$\begin{gathered} C_1(8)=\left(\genfrac{}{}{0ex}{}{8}{1}\right)=\frac{8!}{1!(8-1)!}=\frac{8}{1}=8 \\ {C}_2(2)=\left(\genfrac{}{}{0ex}{}{2}{2}\right)=\frac{2!}{2!(2-2)!}=\frac{1}{1}=1 \\ b=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{2}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{8\cdot 1}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0,0667 \end{gathered}$$

$$\begin{gathered} C_3(8)=\left(\genfrac{}{}{0ex}{}{8}{3}\right)=\frac{8!}{3!(8-3)!}=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=56 \\ c=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{56}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0,4667 \end{gathered}$$

$$\begin{gathered} 3>n_2 \\ d=0 \end{gathered}$$

$e=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{1}\right)+\left(\genfrac{}{}{0ex}{}{n_1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{28\cdot 2+56}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0,9333$

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