Question

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 7 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 7 workers has the same chance of being selected as does any other group (drawing 7 slips without replacement from among 45).
1. How many selections result in all 7 workers coming from the day shift?
2. What is the probability that all 7 selected workers will be from the day shift?
3. What is the probability that all 7 selected workers will be from the same shift?
4. What is the probability that at least two different shifts will be represented among the selected workers?
5. What is the probability that at least one of the shifts will be un-represented in that sample of workers?

Answer 1

Answer:

1. 77520

2. $P_1$ = 0.0017

3. $P_2$ = 0.0019

4. $P_3$ = 0.9981

5. $P_4$ = 0.2036

Step-by-step explanation:

The number of ways or combinations in which we can select x elements from a group of n can be calculated as:

$nCx = \frac{n!}{x!(n-x)!}$

So, there are 77520 selections that result in all 7 workers coming from the day shift. It is calculated as:

$20C7 = \frac{20!}{7!(20-7)!}=77520$

At the same way, the total number of selections of 7 workers from the 45 is 45C7, so the probability that all 7 selected workers will be from the day shift is:

$P_1=\frac{20C7}{45C7} =0.0017$

The probability that all 7 selected workers will be from the same shift is calculated as:

$P_2=\frac{20C7+15C7+10C7}{45C7} =0.0019$

Because the consultant can select all workers from the day shift (20C7) or can select all workers from the swing shift (15C7) or can select all workers from the graveyard shift (10C7).

On the other hand, the probability that at least two different shifts will be represented among the selected workers is the complement of the probability that all 7 selected workers will be from the same shift. So it is calculated as:

$P_3 = 1- P_2=1 - 0.0019 = 0.9981$

Finally, the probability that at least one of the shifts will be un-represented in that sample of workers is:

$P_4=\frac{25C7+30C7+35C7}{45C7} =0.2036$

Where 25C7 is the number of ways to select all 7 workers from swing or graveyard shift, 30C7 is the number of ways to select all 7 workers from day or graveyard shift and 35C7 is the number of ways to selects all 7 workers from day shift and swing shift.