Question

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 4 workers has the same chance of being selected as does any other group (drawing 4 slips without replacement from among 24).

Answer 1

The answer is below

Step-by-step explanation:

A What is the probability that all 4 selected workers will be the day shift?

B What is the probability that all 4 selected workers will be the same shift?

C What is the probability that at least two different shifts will be represented among the selected workers.

A)

The total number of workers = 10 + 8 + 6 = 24

The probability that all 4 selected workers will be the day shift is given as:

`P_a=(C(10,4))/(C(24,4))= (210)/(10626)=0.0198`

`C(n,r)=(n!)/((n-r)!r!)`

B) The probability that all 4 selected workers will be the same shift (
`P_B`) = probability that all 4 selected workers will be the day shift + probability that all 4 selected workers will be the swing shift + probability that all 4 selected workers will be the graveyard shift.

Hence:

`P_B=(C(10,4))/(C(24,4))+(C(8,4))/(C(24,4))+(C(6,4))/(C(24,4))=0.0198+0.0066+0.0014=0.0278`

C) The probability that at least two different shifts will be represented among the selected workers (
`P_C`)= 1 - the probability that all 4 selected workers will be the same shift(
`P_B`)

`P_C=1-P_B\\\\P_C=1-0.0278\\\\P_C=0.972`