1. A Small manufacturing company will start opening a night shift. There are 20

Question

1. A Small manufacturing company will start opening a night shift. There are 20 machinists employed by the company.

a.If the night crew consists of 3 machinists, how many different crews are possible?

b.If the machinists are ranked 1, 2, 3,… ,20 in order of competence, how many of these crews would not have the best machinist?

c.How many of these crews would have at least 1 of the 10 best machinists?

d.If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

e.If If one of these crews is selected at random to work on a particular night, what is the probability that at least two of the worst five machinists work that night?

Explanation / Answer

a. number of different crews possible = 20C3 = 20!/(17!*3!) = 1140

b. We need to find number of crews possible with 19 mechinists (excluding the best one - rank =1) = 19C3 = 969

c. number of crews with at least 1 of the best 10 machinists = Total number of crews - crews without anyone from best 10

= 1140 - 10C3 = 1140 - 120 = 1020

d. Use part a and b for this question,

P(the best machinist will not work that night) = 969/1140 = 0.85

e.

probability that at least two of the worst five machinists work that night = 1- P(one of the worst five works) - P(none of the worst five work) = 1 - 15C2/1140 - 15C3/1140 = 1 - 0.0921- 0.3991 = 0.5088

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