f these workers for in-depth interviews. Suppose the selection is made in such a

Question

f these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 3 A production fadility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shitt. A quality control consultant is to select 3 of workers has the same chance of being selected as does any other group (drawing 3 slips without replacement from among 24) (a) How many selections result in a 3 workers coming from the day shift selections What is the probability that all 3 selected workers will be from the day shift? (Round your answer to four decimal places.) (b) whst is the probability that all 3 selected workers will be from the same shit? (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 fourd (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.) Need Holp?de

Explanation / Answer

a) 10C3 = 120ways

Prob = 10C3/24C3 =120/2024 = 0.0593

b) Prob =10C3/24C3+8C3/24C3 +6C3/24C3 = (120+56+20) /2024 = 196/2024 = 0.0968

c) =1 - 0.0968 = 0.9032

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

P(A1) = 14C3/24C3

P(A2) = 16C3/24C3

P(A3)= 18C3 /24C3

P(A1A2) = 6C3/24C3

P(A1A3) =8C3/24C3

P(A3A2) =10C3/24C3

P(A1A2A3) =0

Prob =(364+ 560+ 816+20+56+120)/2024 = 1936/2024 = 0.9565

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