Probability 66424

There are 5 chocolate, 3 cottage cheese, and 2 apricot croissants in the bag. Croissants are randomly selected in bags. What is the probability of drawing 1 chocolate, 1 cheese, and 1 apricot croissant without returning?

Correct answer:

p =  0.25

Step-by-step explanation:

$$\begin{gathered} C_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ {C}_1(5)=\left(\genfrac{}{}{0ex}{}{5}{1}\right)=\frac{5!}{1!(5-1)!}=\frac{5}{1}=5 \\ {C}_1(3)=\left(\genfrac{}{}{0ex}{}{3}{1}\right)=\frac{3!}{1!(3-1)!}=\frac{3}{1}=3 \\ {C}_1(2)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=\frac{2!}{1!(2-1)!}=\frac{2}{1}=2 \\ p=\frac{\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{5+3+2}{1+1+1}\right)}=\frac{5\cdot 3\cdot 2}{\left(\genfrac{}{}{0ex}{}{5+3+2}{1+1+1}\right)}=\frac{1}{4}=0.25 \end{gathered}$$

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