Question 12.10E3: Employment Assignments A temporary employment agency has six......
Employment Assignments
A temporary employment agency has six men and five women who wish to be assigned for the day. One employer has requested four employees for security guard positions, and the second employer has requested three employees for moving furniture in an office building. If we assume that each of the potential employees has the same chance of being selected and being assigned at random and that only seven employees will be assigned, find the probability that
a) three men will be selected for moving furniture.
b) three men will be selected for moving furniture and four women will be selected for security guard positions.
a) P \left( \begin{matrix} \text{3 men selected}\\ \text{for moving furniture}\end{matrix} \right) = \frac{\left( \begin{matrix} \text{number of possible combinations}\\ \text{of 3 men selected}\end{matrix} \right)}{\left( \begin{matrix} \text{total number of possible combinations}\\ \text{for selecting 3 people}\end{matrix} \right)}
The number of possible combinations with 3 men is _6C_3. The total number of possible selections of 3 people is _{11}C_3.
P \left( \begin{matrix} \text{3 men selected}\\ \text{for moving furniture}\end{matrix} \right) = \frac{_6C_3}{_{11}C_3} = \frac{20}{165} = \frac{4}{33}
Thus, the probability that 3 men are selected is \frac{4}{33}.
b) The number of ways of selecting 3 men out of 6 is _6C_3, and the number of ways of selecting 4 women out of 5 is _5C_4. The total number of possible selections when 7 people are selected from 11 is _{11}C_7. Since both the 3 men and the 4 women must be selected, the probability is calculated as follows:
\begin{aligned} &P \left( \begin{matrix} \text{3 men selected}\\ \text{4 women selected}\end{matrix} \right) = \frac{\left( \begin{matrix} \text{number of combinations}\\ \text{of 3 men selected}\end{matrix} \right) ⋅ \left( \begin{matrix} \text{number of combinations}\\ \text{of 4 women selected}\end{matrix} \right)}{\left( \begin{matrix} \text{total number of possible combinations}\\ \text{for selecting 7 people}\end{matrix} \right)} \end{aligned}
= \frac{_6C_3 ⋅ _5C_4}{_{11}C_7} = \frac{20 ⋅ 5}{330} = \frac{100}{330} = \frac{10}{33}
Thus, the probability is \frac{10}{33}.