Answer:
(a) 210
(b) 0.0018
(c) 0.9982
(d) 0.2180
Explanation:
You want various probabilities related to selecting 6 workers from 24, given 10 work days, 8 work swing, and 6 work the graveyard shift.
(a) All work days
The number of ways that all 6 interviewees can be chosen from the 10 working day shift is ...
n(days) = 10C6 = 210
210 selections result in all 6 being day shift workers.
(b) P(all same shift)
Similarly, the number of selections having all from swing shift is ...
n(swing) = 8C6 = 28
and the number of selections where all are from graveyard is ...
n(grave) = 6C6 = 1
The total number of ways all interviewees can come from the same shift is 210 +28 +1 = 239. Then the probability all are from the same shift is ...
P(all same shift) = (210 +28 +1)/134596 ≈ 0.0018
The probability all are from the same shift is about 0.0018.
(c) P(different)
The probability that the interviewees are from more than one shift is the complement of the probability they are all from one shift:
P(different shifts) = 1 - P(same shift) = 1 - 0.0018 = 0.9982
The probability 2 or more shifts are represented is about 0.9982.
(d) P(1 or 2)
We can count the ways that pairs of shifts are represented. This count will include the count of ways that only one of the two shifts is represented, so we can subtract that out:
n(days & swing only) = (10+8)C6 -10C6 -8C6 = 18564 -210 -28 = 18326
n(days & grave only) = (10+6)C6 -10C6 -6C6 = 8008 -210 -1 = 7797
n(swing & grave only) = (8+6)C6 -8C6 -6C6 = 3003 -28 -1 = 2974
The sum of these is ...
n(2 shifts only) = 18326 +7797 +2974 = 29097
We already know there are 239 ways that only one shift is represented, so we have ...
P(1 or 2 shifts represented) = (239 +29097)/134596 ≈ 0.2180
The probability that at least one shift is unrepresented is about 0.2180.
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