- Aug 10, 2001
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Suppose that there are n people in line waiting to be served sequentially. Each customer, however, will only wait in a line an exponentially distributed amount of time with rate ?. In addition, the service times of each customer is an exponential randon variables with rate µ. Assuming independce among the departure and service times, and assuming someone not in line is currently be served, what is the probability that the nth person will be served?
The answer is P(n) = ((n-1)*? +µ)/(n*? + µ)*P(n-1) which can be solved recursively.
The answer effectively says that the probability that the nth person with be served is the probablly that the (n-1)th will be served times the probability that the nth person doesn't have the smallest departure time. If he has the second smallest departure time, he'll be served? I don't get it.
The answer is P(n) = ((n-1)*? +µ)/(n*? + µ)*P(n-1) which can be solved recursively.
The answer effectively says that the probability that the nth person with be served is the probablly that the (n-1)th will be served times the probability that the nth person doesn't have the smallest departure time. If he has the second smallest departure time, he'll be served? I don't get it.
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