Boys and girls

There are eight boys and nine girls in the class. There were six children on the trip from this class. What is the probability that:

a) only boys went on the field trip
b) just 2 boys went on the field trip

Correct answer:

p₁ =  0.2262 %
p₂ =  28.5068 %

Step-by-step explanation:

$$\begin{gathered} C_6(17)=\left(\genfrac{}{}{0ex}{}{17}{6}\right)=\frac{17!}{6!(17-6)!}=\frac{17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=12376 \\ {C}_6(8)=\left(\genfrac{}{}{0ex}{}{8}{6}\right)=\frac{8!}{6!(8-6)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {p}_1=100\cdot \frac{\left(\genfrac{}{}{0ex}{}{8}{6}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=100\cdot \frac{28}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=0.2262\% \end{gathered}$$

$$\begin{gathered} C_4(9)=\left(\genfrac{}{}{0ex}{}{9}{4}\right)=\frac{9!}{4!(9-4)!}=\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}=126 \\ {C}_2(8)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{2!(8-2)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {p}_2=100\cdot \frac{\left(\genfrac{}{}{0ex}{}{8}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{6-2}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=100\cdot \frac{28\cdot \left(\genfrac{}{}{0ex}{}{9}{6-2}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=28.5068\% \end{gathered}$$

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