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Chapter 5: Problem 57

Events occur according to a Poisson process with rate \(\lambda=2\) per hour. (a) What is the probability that no event occurs between \(8 \mathrm{P} . \mathrm{M} .\) and \(9 \mathrm{P.M.?}\) (b) Starting at noon, what is the expected time at which the fourth event occurs? (c) What is the probability that two or more events occur between \(6 \mathrm{P.M}\). and 8 P.M.?

Short Answer

Expert verified
(a) The probability of no events occurring between 8 PM and 9 PM is \(P(0) = e^{-2}\). (b) The expected time for the fourth event to occur is 2 PM. (c) The probability of two or more events occurring between 6 PM and 8 PM is \(P(k \geq 2) = 1 - (P(0) + P(1))\).

Step by step solution

01

a) Probability of no events occurring between 8 PM and 9 PM

We want to find the probability of no events occurring in a 1-hour interval. In this case, the Poisson PMF can be used, which is given by the formula: \(P(k) = \frac{(\lambda t)^k}{k!}\) \(e^{ -\lambda t}\) where \(k\) is the number of events, \(\lambda\) is the event rate, and \(t\) is the time interval. For this question, \(k = 0\), \(\lambda = 2\) events per hour, and \(t = 1\) hour. Plugging in these values, we get: \(P(0) = \frac{(2 \times 1)^0}{0!}\) \(e^{ -2 \times 1}\) \(P(0) = e^{-2}\) Now calculate the value of \(P(0)\) to find the probability.
02

b) Expected time at which the fourth event occurs

We need to find the expected waiting time for the fourth event to occur starting at noon. For a Poisson process, the waiting times for the events are exponentially distributed with parameter \(\lambda\). The expected waiting time for the \(k\)-th event is given by the formula: \(E[T] = \frac{k}{\lambda}\) In this case, we are interested in the fourth event, so \(k = 4\), and the event rate, \(\lambda = 2\) events per hour. Plugging in these values, we get: \(E[T] = \frac{4}{2}\) \(E[T] = 2\) So, the expected waiting time for the fourth event to occur is 2 hours. As we start at noon, the expected time for the fourth event is 2 hours past noon, which means 2 PM.
03

c) Probability of two or more events occurring between 6 PM and 8 PM

We need to find the probability that two or more events occur in the 2-hour interval between 6 PM and 8 PM. We can do this by calculating the probability of 0 or 1 event occurring in this interval, and then subtracting this probability from 1. We will use the Poisson PMF again. Since the time interval is 2 hours, the rate during this interval will be \(\lambda = 2 \times 2 = 4\). The probability of \(k\) events occurring in this interval is: \(P(k) = \frac{(\lambda t)^k}{k!}\) \(e^{ -\lambda t}\) The probability of 0 or 1 event is the sum of their probabilities, which can be calculated as follows: \(P(0) + P(1) = \frac{(4)^0}{0!}\) \(e^{-4} + \frac{(4)^1}{1!}\) \(e^{-4}\) Now calculate the probability of two or more events occurring between 6 PM and 8 PM: \(P(k \geq 2) = 1 - (P(0) + P(1))\) Plug in the calculated probabilities of 0 and 1 event and compute the final probability value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function (PMF)
In a Poisson process, the probability mass function (PMF) provides a way to calculate the likelihood of a certain number of events happening within a fixed period of time. This function is particularly useful when events occur independently and at a constant average rate. In our given problem, the rate, denoted by \(\lambda\), is given as 2 events per hour.

The formula for the Poisson PMF is:
  • \(P(k) = \frac{(\lambda t)^k}{k!} e^{-\lambda t}\)
where \(k\) is the number of events, \(\lambda\) is the average rate of occurrences, and \(t\) is the time period being considered.

In solving part (a) of the problem, we substituted \(k = 0\), \(\lambda = 2\), and \(t = 1\) hour into this formula to find the probability that no events occur between 8 PM and 9 PM:
Exponentially Distributed Waiting Times
One fascinating aspect of the Poisson process is that the waiting times between events are exponentially distributed. This means the time until the next event follows an exponential distribution, which is characterized by parameter \(\lambda\).

For waiting time problems like part (b) in the exercise, this distribution aids in predicting when a particular event, such as the fourth event, will likely occur. The expected waiting time for the \(k\)-th event is:
  • \(E[T] = \frac{k}{\lambda}\)
In the exercise, we calculated \(E[T] = \frac{4}{2} = 2\), meaning that, starting at noon, we expect the fourth event to occur approximately 2 hours later, at 2 PM.
Expected Value of Events
The expected value plays a pivotal role in understanding the average outcome of random processes, like Poisson processes. When dealing with a sequence of such random events, the expected value tells us what average number of occurrences can be anticipated over a given time frame.

For our problem, part (b) uses the concept to determine the expected time for the fourth event considering the characteristics of our Poisson process. By setting the expected time formula \(E[T] = \frac{k}{\lambda}\), we computed that the fourth event's expected time is 2 PM, considering a start at noon. This reflects the regularity observed in Poisson processes where prediction of event timings improves as events increase.
Probability Calculations
Understanding how to perform probability calculations in a Poisson process is essential for accurately determining outcomes in real-life scenarios. For the exercise part (c), we needed the probability that two or more events occur between 6 PM and 8 PM.

We calculated the probability of 0 or 1 event initially and subtracted these from 1 to find the probability of two or more events:
  • Use the PMF for 0 and 1 event: \(P(0) + P(1)\)
  • Calculate: \(P(2 \text{ or more}) = 1 - (P(0) + P(1))\)
This process shows how complementary probability is utilized to find the likelihood of an event of interest occurring in Poisson processes.

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Most popular questions from this chapter

The lifetimes of A's dog and cat are independent exponential random variables with respective rates \(\lambda_{d}\) and \(\lambda_{c} .\) One of them has just died. Find the expected additional lifetime of the other pet.

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu .\) Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \text { and } X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(E\left[X_{(1)}\right]\) (b) \(\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[X_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

Each entering customer must be served first by server 1 , then by server 2 , and finally by server \(3 .\) The amount of time it takes to be served by server \(i\) is an exponential random variable with rate \(\mu_{i}, i=1,2,3 .\) Suppose you enter the system when it contains a single customer who is being served by server \(3 .\) (a) Find the probability that server 3 will still be busy when you move over to server 2 . (b) Find the probability that server 3 will still be busy when you move over to server 3 . (c) Find the expected amount of time that you spend in the system. (Whenever you encounter a busy server, you must wait for the service in progress to end before you can enter service.) (d) Suppose that you enter the system when it contains a single customer who is being served by server \(2 .\) Find the expected amount of time that you spend in the system.

Consider a two-server parallel queuing system where customers arrive according to a Poisson process with rate \(\lambda\), and where the service times are exponential with rate \mu. Moreover, suppose that arrivals finding both servers busy immediately depart without receiving any service (such a customer is said to be lost), whereas those finding at least one free server immediately enter service and then depart when their service is completed. (a) If both servers are presently busy, find the expected time until the next customer enters the system. (b) Starting empty, find the expected time until both servers are busy. (c) Find the expected time between two successive lost customers.

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). For \(sN(s))\) (b) \(P(N(s)=0, N(t)=3)\); (c) \(E[N(t) \mid N(s)=4] ;\) (d) \(E[N(s) \mid N(t)=4]\).
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