# Exponential Distribution and Expected Times

Suppose two people A,B are assigned to do an individual task and then a group task.

Person A completes his individual task on average around 30 minutes. Person B completes his individual task on average around 40 minutes. After a person completes his individual task, he will move on to the group task. The group task can be completed alone by either A or B on average around 1 hour. The group task can be completed together on average around 30 minutes.

What is the expected time to finish all tasks?

My attempt: I know that it should be a max time it takes for the individual tasks to be done + additional time to finish group task.

I would have

E(max(A,B)) + E(T) where A is time it takes A for his own task and B is time it takes for his own task. The trouble I am having is conditioning on T, which is additional time to do the group task after all individual tasks have been finished.

I know that if can be broken up into two cases and further into two mini-cases:

Case 1: A>B Our time will be A if A > B + T. Our time will be A + T if A < B + T.

Case 2: B>A Our time will be B if A > B + T. Our time will be B + T if A < B + T.

I am not sure how to represent E(T) from here. Intuitively, seems like E(T) could be (0 + Average Time it Takes to Complete Group Task Together) divided by 2, so E(T) = 15 minutes.

Hence my answer would be E(max(A,B)) + E(T) = E(A+B-min(A,B)) + E(T) = 30 + 40 - 120/7 + 15 = 475/7 minutes.

Is my reasoning correct?

It depends on the distribution of times. You have used the words "Exponential distribution" in the title, so let's suppose that the times are distributed that way and use the memoryless property.

Person A works at a rate of $2$ per hour and person B at a rate of $1.5$ per hour on their individual tasks. Individually they work on the group task at a rate of $1$ per hour and together at a rate of $2$ per hour.

• $E[\min(A,B)]=\frac{1}{2+1.5}=\frac{2}{7}$ hours is the expected time to the first task to be completed.

• With probability $\frac{2}{2+1.5}=\frac{4}{7}$ person A finishes the individual task first. If this happens then the conditional additional expected time for the second task is $\frac{1}{1+1.5}=\frac{2}{5}$ hours
• With conditional probability $\frac{1}{1+1.5}=\frac{2}{5}$ person A finishes the group task before B finishes the individual task and person B takes an additional expected $\frac{1}{1.5}=\frac{2}{3}$ hours to finish, with an overall probability $\frac{4}{7}\times \frac{2}{5}=\frac{8}{35}$ and an overall expected time of $\frac{2}{7}+\frac{2}{5}+\frac{2}{3}=\frac{142}{105}$ hours
• With conditional probability $\frac{1.5}{1+1.5}=\frac{3}{5}$ B finishes the individual task before person A finishes the group task and together they take an additional expected $\frac{1}{2}$ hours to finish the group task, with an overall probability $\frac{4}{7}\times \frac{3}{5}=\frac{12}{35}$ and an overall expected time of $\frac{2}{7}+\frac{2}{5}+\frac{1}{2}=\frac{83}{70}$ hours
• With probability $\frac{1.5}{2+1.5}=\frac{3}{7}$ person B finishes the individual task first. If this happens then the conditional additional expected time for the second task is $\frac{1}{1+2}=\frac{1}{3}$ hours

• With conditional probability $\frac{1}{2+1}=\frac{1}{3}$ person B finishes the group task before A finishes the individual task and person A takes an additional expected $\frac{1}{2}$ hours to finish, with an overall probability $\frac{3}{7}\times \frac{1}{3}=\frac{1}{7}$ and an overall expected time of $\frac{2}{7}+\frac{1}{3}+\frac{1}{2}=\frac{47}{42}$ hours
• With conditional probability $\frac{2}{2+1}=\frac{2}{3}$ A finishes the individual task before person B finishes the group task and together they take an additional expected $\frac{1}{2}$ hours to finish the group task, with an overall probability $\frac{3}{7}\times \frac{2}{3}=\frac{2}{7}$ and an overall expected time of $\frac{2}{7}+\frac{1}{3}+\frac{1}{2}=\frac{47}{42}$ hours

As a check, the overall probabilities add up to $1=\frac{8}{35}+\frac{12}{35}+\frac{1}{7}+\frac{2}{7}$.

So the overall expected time is $\frac{8}{35}\times\frac{142}{105} +\frac{12}{35}\times \frac{83}{70}+\frac{1}{7}\times \frac{47}{42}+\frac{2}{7}\times \frac{47}{42} = \frac{251}{210}$ hours. This is about $71$ minutes and $43$ seconds, slightly more than your result.

A different distribution would produce a different result. If you simply thought there were $\frac{1}{2}+\frac{2}{3}+1=\frac{13}{6}$ hours work or $30+40+60=130$ minutes work with no randomness, and two people to do the work, then you would get a result of $\frac{13}{12}$ hours or $65$ minutes required to have it done.

• Thanks. I see it now, but is there any reason why I cannot just have a expectation of max(A,B) + expectation of additional time? The numbers do not equate when I attempted it: Case 1: A<B and A+T < B (8/35)* 0 additional expected time. Case 2: A<B and A+T > B (12/35)*(1/2) additional expected time. Case 3: A>B and A>B+T (1/7) * 0 = 0 additional time. Case 4: A>B and A<B+T (5/7)*(1/2) additional expected time. So E(T) = (5/7)(30) + (12/35)(30) = 222/7 additional minutes to complete all chores. Am I double counting in a way? – Mid Mar 13 '15 at 8:45
• With your approach, I think your case $4$ should use a probability of $\dfrac{2}{7}$ rather than $\dfrac{5}{7}$, giving additional expected time of $\dfrac{11}{35}$ hours or $\dfrac{132}{7}$ minutes. Add this to $E[\max(A,B)]=\dfrac{37}{42}$ hours or $\dfrac{370}{7}$ minutes to give a total expected time of $\dfrac{251}{210}$ hours or $\dfrac{502}{7}$ minutes. – Henry Mar 13 '15 at 9:17
• Oh so the methods are essentially one and the same! Thanks for the help. – Mid Mar 13 '15 at 9:26