Question

Consider a post office that is staffed by two clerks. Suppose that when Miss. Granger enters the system, she discovers that Mr. Harry is being served by one of the clerks and Mr. Ron by other.
Suppose also that Miss Granger is told that her service will begin as soon as either Harry or Ron leaves.
If the amount of time that clerk spends with a customer is exponentially distributed with parameter λ, what is the probability that, of the three customers, Miss. Granger is the last to leave the post office?

1.   1
2.   1/2
3.   1/8
4.   1/4

Answer 1

Good question! It uses the concept of memorylessness in statistics and probability.

Let's go slowly and collect facts and analyze. ( IMHO, collecting clear facts from a probability question gives us 50% confidence. )
 

1. Let $x$ be the time that the first person leaves the system.
2. Let $Y$ be the total waiting time that the remaining person has (starting at time 0)
3. Let $Z$ be the total waiting time that Miss. Granger has (starting at time $x$ ). 

Then the probability that Miss. Granger is not the last person to leave the post office is : 

$P(Y \geq Z + x | Y \geq x )$

The memoryless property of the exponential says that, if X is exponentially distributed, then, 

$P(X \geq x + y | X \geq x) = P(X \geq y)$ 

Substituting X = Y, y = Z, into the above equation we find that, 

$P(Y ≥ Z + x | Y \geq x) = P(Y \geq Z)$

Now, Y and Z are both exponentially distributed random variables with parameter $ \lambda $, so $ P(Y \geq Z) = \frac{1}{2} $. (you can also verify this by symmetry.)

Finally, the answer is $ \frac{1}{2} $  

-sudoankit

Comments

"However, by the memoryless property of exponential random variables, it follows that the additional amount of time that this other person (either Harry or Ron) would still have to spend in the post office"

here,which other person you are talking about.the person whose service is complete and he has gone or the person whose service is gng on..??

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what i ahve concluded from your explanantion is:

if suppose harry leaves first then Ron's service is still gng on and granger can start her servce but due to memoryless property of this distrubution,it will assume that service of ROn is also getting started at the same time when granger's is gettign started.as the service time is equaly likely for both.

hence proba for either of Ron or granger leaving last or first is 1/2

am i correct??

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Yes, you are. For more clarity read this and this lecture notes. They are great.