Consider a post office with two clerks. John, Paul, and Naomi enter simultaneously. John and Paul go directly to the clerks, while Naomi must wait until either John or Paul is finished before she begins service.

MISA6zh

MISA6zh

Answered question

2022-11-06

Consider a post office with two clerks. John, Paul, and Naomi enter simultaneously. John and Paul go directly to the clerks, while Naomi must wait until either John or Paul is finished before she begins service.
(a) If all of the service times are independent exponentially distributed random variables with the same mean 1/lambda, what is the probability that Naomi is still in the post office after the other two have left?
What if we change the assumption in (a) that the two clerks have different service rates lambda_1 and lambda_2?

Answer & Explanation

Kalmukujobvg

Kalmukujobvg

Beginner2022-11-07Added 14 answers

Step 1
Note that Naomi always goes to the clerk that finishes first. Consequently, the behavior of Naomi's random service time is dependent on whether she goes to the first or second clerk (which we will assume to have service rates of λ 1 and λ 2 , respectively). To this end, consider the joint distribution of the random service times of John and Paul:
f X , Y ( x , y ) = f X ( x ) f Y ( y ) = λ 1 λ 2 e ( λ 1 x + λ 2 y ) , x , y > 0 ,
where we have assumed without loss of generality that John went to clerk 1, and Paul went to clerk 2. Then the probability that Naomi is still waiting when both have left is
Pr [ Z > | X Y | ] = Pr [ ( Z > X Y ) ( X > Y ) ] + Pr [ ( Z > Y X ) ( X Y ) ] .
Considering the first term, we have
Pr [ ( Z > X Y ) ( X > Y ) ] = y = 0 x = y ( 1 F Y ( x y ) ) f X , Y ( x , y ) d x d y .
This is because the service time Z follows the distribution of whichever clerk finished first, and in this case, it was Y. I leave it to you to show that this integral is equal to
λ 1 λ 2 ( λ 1 + λ 2 ) 2 ,
and consequently by symmetry,
Pr [ Z > | X Y | ] = 2 λ 1 λ 2 ( λ 1 + λ 2 ) 2 .

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