How many selections result in all 7 workers coming from the day shift?
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 6 of these workers for in-depth interviews.
Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45).
(a) What is the probability that all 6 selected workers will be from the day shift?
(b) What is the probability that all 6 selected workers will be from the same shift?
(c) What is the probability that at least two different shifts will be represented among the selected workers?
(d) What is the probability that at least one of the shifts will be unrepresented in that sample of workers?

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 6 of these workers for in-depth interviews.

Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45).

We will make combinations because we are selecting

part a

P ( DS = 6 ) = 20C6 / 45C6 = 38760 / 8145060 = 0.0047587

part b

P (ALL SAME SHIFT) = P (DS = 6) + P (SS = 6) + P (GS = 6)

= 0.0047587 + (15C6/45C6) + (10C6/45C6)

= 0.0047587 + (5005/8145060) + (210/8145060)

= 0.0047587 + 0.0006145 + 0.000025782 = 0.00539898

part c

The opposite event of the event that at least two different shifts will be represented among the selected workers is the event in B.

Hence the probability is:

P_3 = 1 - P_2 = 1 - 0.00539898 = 0.9946

part d

Suppose that the event that only day shifts be unrepresented is A1, only swing shifts be unrepresented is A2, only graveyard shifts be unrepresented is A3.

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

P (A1 U A2 U A3) = P (A1) + P (A2) + P(A3) - P(A1 & A2) - P(A1 & A3) - P(A2 & A3) + P(A1 & A2 & A3)

where,

P(A1) = 25C6 / 45C6 = 0.02174324069

P(A2) = 30C6 / 45C6 = 0.07290001547

P(A3) = 35C6 / 45C6 = 0.1992815277

P(A1 & A2) = 10C6 / 45C6 = 0.00002578249884

P(A2 & A3) = 20C6 / 45C6 = 0.0047587

P(A1 & A3) = 15C6 / 45C6 = 0.0006145

P(A1 & A2 & A3) = 0

P (A1 U A2 U A3) = 0.02174324069 + 0.07290001547 + 0.1992815277 - 0.00002578249884 - 0.0047587 - 0.0006145 = 0.288526