A production facility employs 10 workers on the day shift, 8 workers on the swin

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 5 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 5 workers has the same chance of being selected as does any other group (drawing 5 slips without replacement from among 24). How many selections result in all 5 workers coming from the day shift? selections What is the probability that all 5 selected workers will be from the day shift? (Round your answer to four decimal places.) What is the probability that all 5 selected workers will be from the same shift? (Round your answer to four decimal places.) What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.) 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.)

Explanation / Answer

There are 24 workers, 10 are from day shift.

A)

There are 10C5 = 10!/(5!5!) = 252 [ANSWER]

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Note that the probability of x successes out of n trials is          
          
P(x) = C(N-K, n-x) C(K, x) / C(N, n)          
          
where          
N = population size =    24      
K = number of successes in the population =    10      
n = sample size =    5      
x = number of successes in the sample =    5      
          
Thus,          
          
P(   5   ) =    0.005928854 [ANSWER]

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b)

There are 10C5 for all day shit, 8C5 for all swing shift, 6C5 from all graveyard shift. Hence,

P = (10C5 + 8C5 + 6C5)/(24C5) = 0.00738754 [ANSWER]

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c)

It is the complement of part b,

P = 1-0.00738754 = 0.99261246 [ANSWER]

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d)

This is the complement of all shifts are represented.

Suppose we get 1 representative from each shift, in 10*8*6/3! = 80 ways. We divide by 3! to reduce the permutations, becasue order does not matter.

Then, we choose randomly in 21C2 ways the other 2.

Hence,

P(all are represented) = 480*(21C2)/(24C5) = 0.395256917

Hence,

P(at least one shift unrepresented) = 1 -0.395256917 = 0.604743083 [ANSWER]

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