Probability question that is driving me nuts:

[gatorpowerpt]

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. :(

 

[mmwave]

Thank you for sharing your question with us. I would approach this problem by first understanding the basic principles of probability. In this scenario, we have a total of 12 employees, of which 3 have been cross-trained to do all 4 tasks. This means that there are a total of 9 employees who can potentially be selected for any given task.

Based on the information provided, it seems that the tasks are randomly assigned to the employees each week. This means that the probability of any employee being selected for a given task is 1/9, or approximately 11.11%. However, as you correctly pointed out, if employee 1 is selected for a task, they cannot be selected for any other tasks that week. This changes the probability for the remaining 8 employees, as there are now only 8 employees available for the remaining tasks.

To calculate the probability of employee 1 being selected for any task, we need to consider all possible outcomes. This can be done through a tree diagram or by using the formula for conditional probability. In this case, the probability of employee 1 being selected for any task would be:

P(employee 1 selected) = P(employee 1 selected for task 1) + P(employee 1 selected for task 2) + P(employee 1 selected for task 3) + P(employee 1 selected for task 4)

= (1/9) + (1/8)*(7/9) + (1/8)*(6/9) + (1/8)*(5/9)

= 0.1975 or approximately 19.75%

This means that there is a 19.75% chance of employee 1 being selected for any task in a given week. However, it is important to note that this is not the same as the probability of employee 1 being selected for a specific task, as this would depend on the specific task and the number of employees available for that task.

I hope this helps to clarify your understanding of the probability in this scenario. If you have any further questions, please do not hesitate to ask. As scientists, it is important for us to always strive for a clear and accurate understanding of concepts and problems. Keep up the good work!