One green

In the container are 45 white and 15 green balls. We randomly select five balls. What is the probability that there will be one green ball maximally?

Correct answer:

p =  0.6329

Step-by-step explanation:

$$\begin{gathered} C_5(60)=\left(\genfrac{}{}{0ex}{}{60}{5}\right)=\frac{60!}{5!(60-5)!}=\frac{60\cdot 59\cdot 58\cdot 57\cdot 56}{5\cdot 4\cdot 3\cdot 2\cdot 1}=5461512 \\ {C}_0(15)=\left(\genfrac{}{}{0ex}{}{15}{0}\right)=\frac{15!}{0!(15-0)!}=\frac{1}{1}=1 \\ {C}_5(45)=\left(\genfrac{}{}{0ex}{}{45}{5}\right)=\frac{45!}{5!(45-5)!}=\frac{45\cdot 44\cdot 43\cdot 42\cdot 41}{5\cdot 4\cdot 3\cdot 2\cdot 1}=1221759 \\ {C}_1(15)=\left(\genfrac{}{}{0ex}{}{15}{1}\right)=\frac{15!}{1!(15-1)!}=\frac{15}{1}=15 \\ {C}_4(45)=\left(\genfrac{}{}{0ex}{}{45}{4}\right)=\frac{45!}{4!(45-4)!}=\frac{45\cdot 44\cdot 43\cdot 42}{4\cdot 3\cdot 2\cdot 1}=148995 \\ a=\left(\genfrac{}{}{0ex}{}{15}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{45}{5}\right)+\left(\genfrac{}{}{0ex}{}{15}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{45}{4}\right)=1\cdot 1221759+15\cdot 148995=3456684 \\ b=\left(\genfrac{}{}{0ex}{}{45+15}{5}\right)=5461512 \\ p=\frac{a}{b}=\frac{3456684}{5461512}=0.6329 \end{gathered}$$

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