A production facility employs 20 workers on the day shift, 15 workers on the swing shift,...

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 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 45).

How many selections result in all 5 workers coming from the day shift?

What is the probability that all 5 selected workers will be from the day shift?

What is the probability that all 5 selected workers will be from the same shift?

What is the probability that at least one of the shifts will be unrepresented in that sample of workers?

What is the probability that at least two different shifts will be represented among the selected workers?

Solutions

Expert Solution

ANSWER:

a) How many selections result in all 5 workers coming from the day shift?

Combinatorial Coefficient (n,k)=n! / k!*(n−k)!

c(20,5) =20!/ (5!(20−5)!) = 15504

What is the probability that all 5 selected workers will be from the day shift?

C(20,5) / C(45,5) = 15504 / 1221759 = 0.0127

(b) What is the probability that all 5 selected workers will be from the same shift?

C(20,5) / C(45,5) +C(15,5) / C(45,5) +C(10,5) / C(45,5)

=15504/1221759+3003/1221759+252/1221759

=0.0127+0.0025+0.0002

=0.0154

P3 = 1- P2

=1-0.0154

=0.9846

(d) What is the probability that at least one of the shifts will be unrepresented in that sample of workers?

c(25,5) / C(45,5)+C(30,5) / C(45,5)+C(35,5)/ C(45,5)-C(10,5)/C(45,5)-C(15,5)/C(45,5)-C(20,5)/C(45,5)+0

=53130/1221759+142506/1221759+324632/1221759-252/1221759-3003/1221759-15504/1221759+0

=0.4105

PLEASE GIVE ME LIKE

orchestra answered 1 year ago

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