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

Answer:

The answer is below

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:

$P_a=\frac{C(10,4)}{C(24,4)}= \frac{210}{10626}=0.0198$

$C(n,r)=\frac{n!}{(n-r)!r!}$

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:

$P_B=\frac{C(10,4)}{C(24,4)}+\frac{C(8,4)}{C(24,4)}+\frac{C(6,4)}{C(24,4)}=0.0198+0.0066+0.0014=0.0278$

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($P_B$)

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