Competition in the class

There are 10 students in the class, of which 8 are girls and two are boys. We want to select three for the competition. What is the probability that they will be:
a) 2 girls and 1 boy
b) 1 girl and 2 boys
c) 3 girls
d) 3 boys
e) at least 2 girls

Correct answer:

a =  0.4667
b =  0.0667
c =  0.4667
d =  0
e =  0.9333

Step-by-step explanation:

$$\begin{gathered} C_2(8)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{2!(8-2)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ {C}_1(2)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=\frac{2!}{1!(2-1)!}=\frac{2}{1}=2 \\ {C}_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ {n}_1=8 \\ {n}_2=2 \\ a=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{28\cdot 2}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0.4667 \end{gathered}$$

$$\begin{gathered} C_1(8)=\left(\genfrac{}{}{0ex}{}{8}{1}\right)=\frac{8!}{1!(8-1)!}=\frac{8}{1}=8 \\ {C}_2(2)=\left(\genfrac{}{}{0ex}{}{2}{2}\right)=\frac{2!}{2!(2-2)!}=\frac{1}{1}=1 \\ b=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{2}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{8\cdot 1}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0.0667 \end{gathered}$$

$$\begin{gathered} C_3(8)=\left(\genfrac{}{}{0ex}{}{8}{3}\right)=\frac{8!}{3!(8-3)!}=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=56 \\ c=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{56}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0.4667 \end{gathered}$$

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

$e=\frac{\left(\genfrac{}{}{0ex}{}{n_1}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{n_2}{1}\right)+\left(\genfrac{}{}{0ex}{}{n_1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n_1+n_2}{3}\right)}=\frac{28\cdot 2+56}{\left(\genfrac{}{}{0ex}{}{8+2}{3}\right)}=0.9333$

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