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Math Help - Exponential expected service time question

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    Exponential expected service time question

    A customer enters a store where he is serviced by clerk 1, followed by clerk 2, finally by clerk 3. Service time by each of the clerks is an exponentially distributed variable with \mu_i, where i =1,2,3. Assume you enter the the store and there is a single customer being serviced by clerk 3.

    a) Find the probability that clerk 3 will still be busy when you move to clerk 2.
    b) Find the probability that clerk 3 will remain busy when you move to see him.
    c) Find the total time in the store

    Solutions:

    a) For this I'm thinking that \frac{1}{\mu_3}>\frac{1}{\mu_1} for clerk 3 to be busy when you finish with clerk 1. Therefore the probability that clerk 3 is still busy when you move to clerk 2 would be P(C_1\leq C_3) = \frac{\mu_1}{\mu_1+\mu_2+\mu_3}. But this doesn't seem correct.

    b) You're done with clerk 1 and 2 and you have to wait to see clerk 3. Thus the expected time of clerk 3 would have to be greater then that of clerk 1 and 2 combined so you would have \frac{1}{\mu_3}>\frac{1}{\mu_1+\mu_2} which would imply P(C_1 \ \cup \ C_2 \leq C_3)= \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}

    c) For this one I'm assuming it would be the total service time + total waiting time (assuming clerk 3 is still busy when you go see him)
    so I would have :

    E[T] = \frac{1}{\mu_1} +\frac{1}{\mu_2} +\frac{1}{\mu_3} +\left( \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}\right) \times \left( \frac{1}{\mu_3}\right)
    Last edited by lllll; November 3rd 2008 at 09:59 PM.
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    Quote Originally Posted by lllll View Post
    A customer enters a store where he is serviced by clerk 1, followed by clerk 2, finally by clerk 3. Service time by each of the clerks is an exponentially distributed variable with \mu_i, where i =1,2,3. Assume you enter the the store and there is a single customer being serviced by clerk 3.

    a) Find the probability that clerk 3 will still be busy when you move to clerk 2.
    b) Find the probability that clerk 3 will remain busy when you move to see him.
    c) Find the total time in the store

    Solutions:

    a) For this I'm thinking that \frac{1}{\mu_3}>\frac{1}{\mu_1} for clerk 3 to be busy when you finish with clerk 1. Therefore the probability that clerk 3 is still busy when you move to clerk 2 would be P(C_1\leq C_3) = \frac{\mu_1}{\mu_1+\mu_2+\mu_3}. But this doesn't seem correct.

    b) You're done with clerk 1 and 2 and you have to wait to see clerk 3. Thus the expected time of clerk 3 would have to be greater then that of clerk 1 and 2 combined so you would have \frac{1}{\mu_3}>\frac{1}{\mu_1+\mu_2} which would imply P(C_1 \ \cup \ C_2 \leq C_3)= \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}

    c) For this one I'm assuming it would be the total service time + total waiting time (assuming clerk 3 is still busy when you go see him)
    so I would have :

    E[T] = \frac{1}{\mu_1} +\frac{1}{\mu_2} +\frac{1}{\mu_3} +\left( \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}\right) \times \left( \frac{1}{\mu_3}\right)
    Assuming service by Clerk 3 to customer begins at the same as service by Clerk 1 to you:

    (a) Calculate \Pr(X_3 - X_1 > 0) = \Pr(X_3 > X_1) = \int_{x_1 = 0}^{x_1 = + \infty} \int_{x_3 = x_1}^{x_3 = + \infty} f(x_1, x_3) \, dx_3 \, dx_1 where f(x_1, x_3) = \left( \frac{1}{\mu_1} e^{-x_1/\mu_1}\right) \left( \frac{1}{\mu_3} e^{-x_3/\mu_3}\right)

    assuming service times by each clerk are independent.


    b) Calculate \Pr(X_3 - X_2 - X_1 > 0).
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