- #1
gatorpowerpt
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I was asked this question about a week ago and I can't figure it out. I get confused in the numbers. This is not a homework question, but I am going to phrase it like one so maybe it would be easier to conceptualize.
Suppose you originally had 12 employees and 4 mundane tasks to do each week. Instead of training 12 people to each do all 4 jobs, you split the employees into groups of three so they would only have to know one job, but rotating it every three weeks. i.e.:
Job 1 - Job 2 - Job 3 - Job 4
employee 1 - employee 4 - employee 7 - employee 10
employee 2 - employee 5 - employee 8 - employee 11
employee 3 - employee 6 - employee 9 - employee 12
Three employees quit and you cross-train employee 1 to do all 4 tasks:
J1 - J2 - J3 - J4
e1 - e1 - e1 - e1
e2 - e5 - e8 - e11
e3 - e6 - e9 - e12
...because of this change, you're no longer going to rotate weeks, but instead use a computer to randomly assign jobs to employees trained to do that work. If employee 1 is selected at any point, they are automatically eliminated from doing any of the other jobs that week (i.e. employee 1 can not be picked to do all four jobs, only one)
Even so, he is more likely to be picked each week. What is this percentage/probability?
So I said to myself, we'll they're each 1/3
J1 - J2 - J3 - J4
e01 (1/3) - e01 (1/3) - e01 (1/3) - e01 (1/3)
e02 (1/3) - e05 (1/3) - e08 (1/3) - e11 (1/3)
e03 (1/3) - e06 (1/3) - e09 (1/3) - e12 (1/3)
but if e01 gets selected then the other changes to 1/2
J1 - J2 - J3 - J4
e01 (1/1) - e01 (0) - e01 (0) - e01 (0)
e02 (0) - e05 (1/2) - e08 (1/2) - e11 (1/2)
e03 (0) - e06 (1/2) - e09 (1/2) - e12 (1/2)
but that doesn't really help me decide how likely e01 would get picked, just how it would change the matrix if they did get picked. So I said, well, what is the chances that it DOES NOT GET PICKED?
J1 - J2 - J3 - J4
e01 (0) - e01 (0) - e01 (0) - e01 (0)
eXX (2/3) - eXX (2/3) - eXX (2/3) - eXX (2/3)
which gives me: 16/81 (19.75%)
Does that mean the likelihood if him getting picked is: 16/81 - 1 (80.24%)
I am not sure how to answer this question. :(
Suppose you originally had 12 employees and 4 mundane tasks to do each week. Instead of training 12 people to each do all 4 jobs, you split the employees into groups of three so they would only have to know one job, but rotating it every three weeks. i.e.:
Job 1 - Job 2 - Job 3 - Job 4
employee 1 - employee 4 - employee 7 - employee 10
employee 2 - employee 5 - employee 8 - employee 11
employee 3 - employee 6 - employee 9 - employee 12
Three employees quit and you cross-train employee 1 to do all 4 tasks:
J1 - J2 - J3 - J4
e1 - e1 - e1 - e1
e2 - e5 - e8 - e11
e3 - e6 - e9 - e12
...because of this change, you're no longer going to rotate weeks, but instead use a computer to randomly assign jobs to employees trained to do that work. If employee 1 is selected at any point, they are automatically eliminated from doing any of the other jobs that week (i.e. employee 1 can not be picked to do all four jobs, only one)
Even so, he is more likely to be picked each week. What is this percentage/probability?
So I said to myself, we'll they're each 1/3
J1 - J2 - J3 - J4
e01 (1/3) - e01 (1/3) - e01 (1/3) - e01 (1/3)
e02 (1/3) - e05 (1/3) - e08 (1/3) - e11 (1/3)
e03 (1/3) - e06 (1/3) - e09 (1/3) - e12 (1/3)
but if e01 gets selected then the other changes to 1/2
J1 - J2 - J3 - J4
e01 (1/1) - e01 (0) - e01 (0) - e01 (0)
e02 (0) - e05 (1/2) - e08 (1/2) - e11 (1/2)
e03 (0) - e06 (1/2) - e09 (1/2) - e12 (1/2)
but that doesn't really help me decide how likely e01 would get picked, just how it would change the matrix if they did get picked. So I said, well, what is the chances that it DOES NOT GET PICKED?
J1 - J2 - J3 - J4
e01 (0) - e01 (0) - e01 (0) - e01 (0)
eXX (2/3) - eXX (2/3) - eXX (2/3) - eXX (2/3)
which gives me: 16/81 (19.75%)
Does that mean the likelihood if him getting picked is: 16/81 - 1 (80.24%)
I am not sure how to answer this question. :(