Probability 38041

Seven women and 3 men work in one office. According to the new regulation, reducing the number of employees by three is necessary. In a random sample of employees, what is the probability that they will be fired:
a. One woman and two men
b. At least one woman

Correct answer:

p₁ =  0.175
p₂ =  0.7

Step-by-step explanation:

$$\begin{gathered} z=7 \\ m=3 \\ {C}_1(3)=\left(\genfrac{}{}{0ex}{}{3}{1}\right)=\frac{3!}{1!(3-1)!}=\frac{3}{1}=3 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \frac{7}{10}\cdot \frac{3}{9}\cdot \frac{2}{8}=3\cdot \frac{7}{10}\cdot \frac{3}{9}\cdot \frac{2}{8}=\frac{7}{40}=0.175 \end{gathered}$$

$p_2=\frac{7}{10}\cdot \frac{9}{9}\cdot \frac{8}{8}=\frac{7}{10}=0.7$

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