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).

(a) 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.)

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

a)  Select ( 5 Day workers ) = 252,  P ( 5 Day Workers ) = 0.006

b) P ( 5 same shift )  = 0.0074612

c)  P ( At-least 2 different shifts ) = 0.9925

d) P ( Only 2 different shifts )  = 0.3366

Step-by-step explanation:

Given:-

- The number of day shift workers D = 10

- The number of swing shift workers S = 8

- The number of graveyard shift workers G = 6

- The total selection made by the Quality Team = 5

Find:-

(a) 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?

Solution:-

- To select the slips such that all 5 are for Day shift workers, in other words you are also selecting 5 Day shift workers from a pool of 10 Day shift workers. The following number of combinations would be:

                          Select ( 5 Day workers ) = 10 C 5 = 252 combinations

- The total possible outcomes for selecting 5 workers from any of the shifts is:

                         Select ( 5 Workers ) = 24 C 5 = 42504 combinations

- The associated probability for selecting 5 day shift workers is:

          P ( 5 Day Workers ) = Select ( 5 Day workers ) /  Select ( 5 Workers )

                                             = 252 / 42504

                                             = 0.006

Find:-

(b) What is the probability that all 5 selected workers will be from the same shift? (Round your answer to four decimal places.)

Solution:-

- To select the slips such that all 5 are from same shift, in other words you are selecting 5 Day shift workers, or 5 Swing shift workers or 5 Graveyard shift workers from a pool of 10 Day shift workers, 8 Swing shift workers, 6 graveyard shift workers. The following number of combinations would be:

                          Select ( 5 Day workers ) = 10 C 5 = 252 combinations

                          Select ( 5 Swing workers ) = 8 C 5 = 56 combinations

                          Select ( 5 Graveyard workers ) = 6 C 5 = 6 combinations

- The associated probability for selecting 5 same shift workers is:

  P ( 5 Day Workers ) = Select ( 5 Day workers ) /  Select ( 5 Workers )

                                             = 252 / 42504

                                             = 0.006

 P ( 5 Swing Workers ) = Select ( 5 swing workers ) /  Select ( 5 Workers )

                                             = 56 / 42504

                                             = 0.00132

P ( 5 Graveyard Workers ) = Select ( 5 graveyard workers ) /  Select ( 5 Workers )

                                             = 6 / 42504

                                             = 0.0001412

- P ( 5 same shift )  = P ( 5 Day Workers ) + P ( 5 Swing Workers ) + P ( 5 Graveyard Workers )

                                = 0.006 + 0.00132 + 0.0001412

                                =  0.0074612

Find:-

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

Solution:-

- To select the slips such that all 5 are from different shifts, in other words you are selecting either a combination of Day shift workers and Swing shift workers or Day shift and Graveyard shift workers or Swing shift and Graveyard shift workers or a combination of all 3. It would be easier if we subtract the probability of no different workers from 1 to get at-least 2 different workers probability. As follows:

- P ( At-least 2 different shifts )  = 1 - P ( 5 same shift Workers )

                                                      = 1 - 0.0074612

                                                      = 0.9925

Find:-

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

Solution:-

- To select the slips such that all 5 are from only 2 different shifts, in other words you are selecting either a combination of Day shift workers and Swing shift workers or Day shift and Graveyard shift workers or Swing shift and Graveyard shift workers but not a combination of all 3. It would be easier if we subtract the probability of all different workers from the probability of at-least 2 different workers. As follows:

- P ( Only 2 different shifts )  = P ( At-least 2 different ) - P ( All 3 shift Workers )

- The possible combinations for all 3 different shift workers is:

Select ( 3 D ,  1 S , 1 G ) = 10 C 3 * 8 * 6 = 5,760

Select ( 2 D ,  2 S , 1 G ) = 10 C 2 * 8 C 2 * 6 = 7,560

Select ( 2 D ,  1 S , 2 G ) = 10 C 2 * 8  * 6 C 2 = 5,400

Select ( 1 D ,  2 S , 2 G ) = 10 * 8 C 2 * 6C2 = 4,200

Select ( 1 D ,  3 S , 1 G ) = 10 * 8 C 3 * 6 = 3,360

Select ( 1 D ,  1 S , 3 G ) = 10 * 8  * 6 C 3 = 1,600

Total All 3 different shifts selected  = 27,880

P ( All 3 shift Workers ) = 27,880 / 42504 = 0.655938264

- Hence,

P ( Only 2 different shifts ) = 0.9925 - 0.655938264

                                                = 0.3366

the f

last graph i believe

step-by-step explanation:

worked it out i believe