# Thread: Exponential expected service time question

1. ## 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 :

$E[T] = \frac{1}{\mu_1} +\frac{1}{\mu_2} +\frac{1}{\mu_3} +\left( \frac{\mu_1+\mu_2}{\mu_1+\mu_2+\mu_3}\right) \times \left( \frac{1}{\mu_3}\right)$

2. Originally Posted by lllll
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 $\Pr(X_3 - X_2 - X_1 > 0)$.