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 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 belng selected as does any other group (drawing 4 slips without replacement from among 24) (a) How many selections result in all 4 workers coming from the day shift? x selections What is the probability that all 4 selected workers will be from the day shift? (Round your answer to four decimal places.) (b) What is the probability that all 4 selected workers wll be from the same shift? (Round your answer to four decimal places.) (c) What is the probablity that at least two different shifts ll 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.)

Explanation / Answer

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We have to employ 10 workers

a. ways in which 4 out of 4 are choosen from day shift

= 10C4 as we are selecting any 4 from the 10 in day shift

= 210

b. This is basically 10C4 + 8C4 + 6C4 / 24C4 = 295/10626 = .0278

c. In this case we will calculate probability that all workers come from same shift. Subtract this number from 1 to get : 1 - ,0278 = .9722

d. P(atleast one of the shifts will be unrepresented) = 1-P(all 3 are represented)

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 A2.

The probability that at least one of the shifts will be unrepresented in the sample of workers is P4 = P(A1A2A3) = P(A1)+P(A2)+P(A3)P(A1A2)P(A1A3)P(A3A2)+P(A1A2A3)

P(A1 U A2 U A3) = 14C4 + 16C4 + 18C4 - 6C4 - 8C4-10C4 + 0 / 24C4

=5586/10626

= 0.5257

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