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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 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 24).

(a) How many selections result in all 6 workers coming from the day shift?
252

Incorrect: Your answer is incorrect.
selections

What is the probability that all 6 selected workers will be from the day shift? (Round your answer to four decimal places.)


(b) What is the probability that all 6 selected workers will be from the same shift? (Round your answer to four decimal places.)


(c) What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.)


(d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers? (Round your answer to four decimal places.)

asked
User Dimme
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1 Answer

5 votes

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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answered
User Gentatsu
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