Chapter 5.4, Problem 35E

Day and night shifts: A company has hired 12 new employees, and must assign 8 to the day shift and 4 to the night shift.

1. In how many ways can the assignment be made?
2. Assume that the 12 employees consist of six men and women and that the assignments to day and night shift are made at random. What is the probability that all four of the night-shift employees are men?
3. What is the probability that at least one of the night-shift employees is a woman?

To determine

To find:The number of ways can the assignment be made.

Answer to Problem 35E

The required number of different assignments is $\underset{\_}{495}.$

Explanation of Solution

Given information:

A company has hired 12 new employees and must assign 8 to the day shift and 4 to the night shift.

Concept Used:

We have the fundamental counting principle which states that if there are three events that can occur in m, n and p ways respectively then the total number of ways in which all three events can occur is $m.n.p$ .

We know that the permutations of objects are the arrangements of the objects. And in the arrangement of objects, the order of objects does matter. For example, if we have three objects say, A, B, and C, there are a total of six arrangements of the objects that are possible and these are; ABC, ACB, BAC, BCA, CAB, and CBA.

We know that combination is a selection of r objects out of n different objects and order o the selection of objects does not matter. For example, if there are three objects say, P, Q, and R and we have to select two objects out of three. Then while selection, selecting P and then Q is the same as selecting Q and then P. So, in other words selecting PQ is the same as selecting QP.

Calculation:

So, according to the question, the number of ways of selecting 8 employees for day shift out of 12 and the remaining 4 employees for night shift out of 4 remaining employees is;

  $\begin{array}{l}\text{Number}\text{of}\text{ways}=C{}_8\times C{}_4\\ =\frac{12!}{8!\left(12-8\right)!}\times \frac{4!}{4!\left(4-4\right)!}\\ =\frac{12!}{8!\left(4\right)!}\times \frac{4!}{4!\left(0\right)!}\\ =495\times 1=495\\ \text{Number}\text{of}\text{ways}=495\end{array}$

The required number of different assignments is $\underset{\_}{495}.$

To determine

To find:The probability that all four of the night-shift employees are men by assuming that the 12 employees consist of six men and six women and that the assignments for day and night shift are made at random.

Answer to Problem 35E

Probability that all 4 of the night-shift employees are men is $\underset{\_}{0.03}.$

Explanation of Solution

Given information:

A company has hired 12 new employees and must assign 8 to the day shift and 4 to the night shift.

Concept Used:

We have the fundamental counting principle which states that if there are three events that can occur in m, n and p ways respectively then the total number of ways in which all three events can occur is $m.n.p$ .

We know that the permutations of objects are the arrangements of the objects. And in the arrangement of objects, the order of objects does matter. For example, if we have three objects say, A, B, and C, there are total six arrangements of the objects are possible and these are; ABC, ACB, BAC, BCA, CAB, and CBA.

We know that combination is a selection of r objects out of n different objects and order o the selection of objects does not matter. For example, if there are three objects say, P, Q, and R and we have to select two objects out of three. Then while selection, selecting P and then Q is the same as selecting Q and then P. So, in other words selecting PQ is the same as selecting QP.

Calculation:

Number of ways of selecting 4 men out of 6 men for night shift and rest employees for the day shift is;

  $\begin{array}{l}C{}_4\times C{}_8=\frac{6!}{4!\left(6-4\right)!}\times \frac{8!}{8!\left(8-8\right)!}\\ =\frac{6!}{4!\left(2\right)!}\times \frac{8!}{8!\left(0\right)!}\\ =15\times 1=15\end{array}$

Hence the required number of different assignments is 15.

So now according to the question, the probability that all 4 of the night-shift employees are men is calculated as follows;

  $\begin{array}{l}\text{Required}\text{probability}=\frac{C{}_4\times C{}_8}{C{}_4}\\ =\frac{15}{495}=0.03\end{array}$

To determine

To find:The probability that at least one of the night-shift employees is a woman.

Answer to Problem 35E

  $\underset{\_}{0.97}.$

Explanation of Solution

Given information:

A company has hired 12 new employees and must assign 8 to the day shift and 4 to the night shift.

Concept Used:

We have the fundamental counting principle which states that if there are three events that can occur in m, n and p ways respectively then the total number of ways in which all three events can occur is $m.n.p$ .

We know that the permutations of objects are the arrangements of the objects. And in the arrangement of objects, the order of objects does matter. For example, if we have three objects say, A, B, and C, there are total six arrangements of the objects are possible and these are; ABC, ACB, BAC, BCA, CAB, and CBA.

We know that combination is a selection of r objects out of n different objects and order o the selection of objects does not matter. For example, if there are three objects say, P, Q, and R and we have to select two objects out of three. Then while selection, selecting P and then Q is the same as selecting Q and then P. So, in other words selecting PQ is the same as selecting QP.

Calculation:

According to the question, the probability that at least one of the night-shift employees is a woman is calculated as follows;

  $\begin{array}{l}P\left(\text{at}\text{least}\text{one}\text{is}\text{a}\text{woman}\right)=1-P\left(\text{No}\text{women}\right)\\ =1-0.03\\ =0.97\\ \end{array}$

Probability is $\underset{\_}{0.97}.$