In class 24

There are a total of 16 students in the class, a quarter of whom are girls. A team of five is selected at random. Determine the probability that the team will have:

a) at least 4 girls,
b) at most 1 girl,
c) no girls.

Final Answer:

p₁ =  0.0027
p₂ =  0.6346
p₃ =  0.1813

Step-by-step explanation:

$$\begin{gathered} C_5(16)=\left(\genfrac{}{}{0ex}{}{16}{5}\right)=\frac{16!}{5!(16-5)!}=\frac{16\cdot 15\cdot 14\cdot 13\cdot 12}{5\cdot 4\cdot 3\cdot 2\cdot 1}=4368 \\ {C}_4(4)=\left(\genfrac{}{}{0ex}{}{4}{4}\right)=\frac{4!}{4!(4-4)!}=\frac{1}{1}=1 \\ {C}_1(12)=\left(\genfrac{}{}{0ex}{}{12}{1}\right)=\frac{12!}{1!(12-1)!}=\frac{12}{1}=12 \\ n=16 \\ d=n/4=16/4=4 \\ c=n-d=16-4=12 \\ T=\left(\genfrac{}{}{0ex}{}{n}{5}\right)=4368 \\ {p}_1=\frac{\left(\genfrac{}{}{0ex}{}{d}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{c}{1}\right)}{T}=\frac{1\cdot 12}{4368}=0.0027 \end{gathered}$$

$$\begin{gathered} C_0(4)=\left(\genfrac{}{}{0ex}{}{4}{0}\right)=\frac{4!}{0!(4-0)!}=\frac{1}{1}=1 \\ {C}_5(12)=\left(\genfrac{}{}{0ex}{}{12}{5}\right)=\frac{12!}{5!(12-5)!}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8}{5\cdot 4\cdot 3\cdot 2\cdot 1}=792 \\ {C}_1(4)=\left(\genfrac{}{}{0ex}{}{4}{1}\right)=\frac{4!}{1!(4-1)!}=\frac{4}{1}=4 \\ {C}_4(12)=\left(\genfrac{}{}{0ex}{}{12}{4}\right)=\frac{12!}{4!(12-4)!}=\frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}=495 \\ {d}_0=\left(\genfrac{}{}{0ex}{}{d}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{c}{5}\right)=1\cdot 792=792 \\ {d}_1=\left(\genfrac{}{}{0ex}{}{d}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{c}{4}\right)=4\cdot 495=1980 \\ {p}_2=\frac{d_0+d_1}{T}=\frac{792+1980}{4368}=0.6346 \end{gathered}$$

$p_3=\frac{d_0}{T}=\frac{792}{4368}=0.1813$

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