Question

Excited by the recent warm weather Jill and Kelly are doing spring cleaning at their apartment. Jill takes an exponentially distributed amount of time with mean 30 minutes to clean the kitchen. Kelly takes an exponentially distributed amount of time with mean 40 minutes to clean the bath room. The first one to complete their task will go outside and start raking leaves, a task that takes an exponentially distributed amount of time with a mean of one hour. When the second person is done inside, they will help the other and raking will be done at rate 2. (Of course the other people may already be done ranking in which case the chores are done.) What is the expected time until the chores are all done?

I saw this problem solved by others, but it wasn't done right. Please show all your work.

Answer 1

Here I am assuming that the statement "When the second person is done inside, they will help the other and raking will be done at rate 2" means jointly they can do it at a mean rate of 60/2=30 mins.

Jill works at a rate of 2 per hour and Kelly 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.

Let A=Time taken by Jill in first task.B:Time taken by Kelly in first task.

So Min(A,B) is the time taken for the first person to be free.

Now P((Min(A,B))>t)=P(A>t)*P(B>t)=exp(-t/30)*exp(-t/40) which is itself an exponential distribution a it is of the form exp(-t*a) where a>0.

E[min(A,B)]=1/(2+1.5)= 2/7 hours is the expected time to the first task to be completed.

Now, With probability 2/(2+1.5)=4/7 person Jill finishes the individual task first. If this happens then the conditional additional expected time for the second task is 1/1(+1.5)=2/5 hours.

Thus with conditional probability 1/(1+1.5)=2/5 Jill finishes the group task before Kelly finishes the individual task and Kelly takes an additional expected 1/1.5=2/3 hours to finish, with an overall probability 4/7×2/5=8/35 and an overall expected time of 2/7+2/5+2/3=142/105 hours.

Now, With conditional probability 1.5/(1+1.5)=3/5 Kelly finishes the individual task before Jill finishes the group task and together they take an additional expected 1/2 hours to finish the group task, with an overall probability 4/7×3/5=12/35 and an overall expected time of 2/7+2/5+1/2=83/70 hours

With probability 1.5/(2+1.5)=3/7 Kelly finishes the individual task first. If this happens then the conditional additional expected time for the second task is 1(1+2)=1/3 hours

Thus,With conditional probability 1/(2+1)=1/3 Kelly finishes the group task before Jill finishes the individual task and Jill takes an additional expected 1/2 hours to finish, with an overall probability 3/7×1/3=1/7 and an overall expected time of 2/7+1/3+1/2=47/42 hours

Now, With conditional probability 2/(2+1)=2/3 Jill finishes the individual task before Kelly finishes the group task and together they take an additional expected 1/2 hours to finish the group task, with an overall probability 3/7×2/3=2/7 and an overall expected time of 2/7+1/3+1/2=47/42 hours

So the overall expected time is 8/35×142/105+12/35×83/70+1/7×47/42+2/7×47/42=251/210 Ans.