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 left

a) only boys
b) just two boys

Correct answer:

p₁ =  0.2262 %
p₂ =  28.5068 %

Step-by-step explanation:

$$\begin{gathered} C_6(17)=\left(\genfrac{}{}{0ex}{}{17}{6}\right)=\frac{17!}{6!(17-6)!}=\frac{17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=12376 \\ {C}_6(8)=\left(\genfrac{}{}{0ex}{}{8}{6}\right)=\frac{8!}{6!(8-6)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {p}_1=100\cdot \frac{\left(\genfrac{}{}{0ex}{}{8}{6}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=100\cdot \frac{28}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=\frac{50}{221}\%=0.2262\% \end{gathered}$$

$$\begin{gathered} C_4(9)=\left(\genfrac{}{}{0ex}{}{9}{4}\right)=\frac{9!}{4!(9-4)!}=\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}=126 \\ {C}_2(8)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{2!(8-2)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {p}_2=100\cdot \frac{\left(\genfrac{}{}{0ex}{}{8}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{6-2}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=100\cdot \frac{28\cdot \left(\genfrac{}{}{0ex}{}{9}{6-2}\right)}{\left(\genfrac{}{}{0ex}{}{8+9}{6}\right)}=\frac{6300}{221}\%=28.5068\% \end{gathered}$$

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