Respuesta :
A)
We will solve using:
[tex]_{22}C_4=\frac{22\cdot21\cdot20\cdot19}{4\cdot3\cdot2}=7315[/tex]The number of ways to choose 4 girls from 12 girls and 0 boys from 10 boys is:
[tex]_{10}C_0\cdot_{12}C_4=1\cdot\frac{12!}{8!\cdot4!}=\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}=495[/tex]So, the probability() of choosing 4 just 4 girls is:
[tex]P=\frac{_{10}C_0\cdot_{12}C_4}{_{22}C_4}\Rightarrow P=\frac{495}{7315}\Rightarrow P=\frac{9}{133}\Rightarrow P\approx0.068[/tex]So, that is the probability to get just 4 girls on the committee.
B)
For the committee to be just 4 boys is found as follows:
[tex]_{22}C_4=\frac{22\cdot21\cdot20\cdot19}{4\cdot3\cdot2}=7315[/tex]And the number of ways to choose 4 boys from 10 boys and 0 girls from 12 girls is:
[tex]_{10}C_4\cdot_{12}C_0=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2}=210[/tex]So, the probability to get just 4 boys on the committee is:
[tex]P=\frac{_{10}C_4\cdot_{12}C_0}{_{22}C_4}\Rightarrow P=\frac{210}{7315}\Rightarrow P=\frac{6}{209}\Rightarrow P\approx0.028[/tex]C)
For the committee to have at least one girl is:
[tex]_{22}C_4=\frac{22\cdot21\cdot20\cdot19}{4\cdot3\cdot2}=7315[/tex]No. of ways to selecting at least 1 girl:
22C4 - 10C4 = 7315 - 210 = 7105
Now, we calculate the probability:
[tex]P=\frac{7105}{7315}\Rightarrow P=\frac{203}{209}\Rightarrow P\approx0.971[/tex]