Suppose patients are coming into a hospital at an average rate of 4 per hour, ac

Question

Suppose patients are coming into a hospital at an average rate of 4 per hour, according to a Poisson arrival process. Denote N(2, 4) to be the number of patients coming in between the 2nd to the 4th hour. Recall that the waiting time between two consecutive patient follows the exponential(4) distribution. Please calculation (a) the probability that the fifth patient comes takes more than 30 minutes to arrive. (b) the probability that the first patient arrives after t = 0 takes less than 1 hour to arrive, and the waiting time between the fourth and fifth patient is more than 3 hours.

Explanation / Answer

(a)

The waiting time between two consecutive patient follows the exponential(4) distribution.

That is, the mean arrival of patients = 4 per hour

or, mean waiting time between two consecutive patient = 1/4 = 0.25 hour

So, mean arrival of 5 patients = 0.25 * 5 = 1.25

The probability that the fifth patient comes takes more than 30 minutes (0.5 hr) to arrive

= P(X > 0.5) = exp(-1.25 * 0.5) = 0.5353

(b)

Mean arrival of patients = 4 per hour

The probability that the first patient arrives after t = 0 takes less than 1 hour to arrive

= P(X < 1) = 1 - exp(-4 * 1) = 0.9817

By memoryless property of exponential distribution, the waiting time between the fourth and fifth patient follows the exponential(4) distribution.

So, the probability that the waiting time between the fourth and fifth patient is more than 3 hours

= P(X > 3) = exp(-4 * 3) = 00000.6

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