/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Events occur according to a Pois... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Events occur according to a Poisson process with rate \(\lambda=2\) per hour. (a) What is the probability that no event occurs between 8 P.M. and 9 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 that no event occurs between 8 P.M. and 9 P.M. is approximately 0.135. (b) The expected time at which the fourth event occurs, starting at noon, is 2 P.M. (c) The probability that two or more events occur between 6 P.M. and 8 P.M. is approximately 0.982.

Step by step solution

01

(a) Probability of no events between 8 P.M. and 9 P.M.

The probability of observing x events in time interval t is given by the Poisson probability function: \(P(X = x) = \frac{e^{-\lambda t}(\lambda t)^x}{x!}\) Here, we want to find the probability that no events occur in a one-hour interval (8 P.M. to 9 P.M.), which means x = 0. So, the probability function for no events occurring (x = 0) is: \(P(X = 0) = \frac{e^{-\lambda}(\lambda)^0}{0!}\) Substituting λ = 2, we obtain: \(P(X = 0) = \frac{e^{-2}(2)^0}{0!} = \frac{e^{-2}(1)}{1} = e^{-2} \approx 0.135\) Hence, the probability that no event occurs between 8 P.M. and 9 P.M. is approximately 0.135.
02

(b) Expected time for the fourth event

In a Poisson process, the time between events follows an exponential distribution. The average time between events, or inter-arrival time, is given by: \(E[T] = \frac{1}{\lambda}\) For this problem, λ = 2 events per hour, so: \(E[T] = \frac{1}{2} = 0.5\) hours Since we are looking for the expected time for the fourth event, we need to consider the time it takes for the first three events to occur as well: Expected time for four events = Expected time for one event × 4 = (0.5 hours) × 4 = 2 hours The problem states that we are starting at noon. Therefore, the expected time for the fourth event to occur would be: Noon + 2 hours = 2 P.M. The expected time at which the fourth event occurs is 2 P.M.
03

(c) Probability of two or more events between 6 P.M. and 8 P.M.

In this case, we are interested in the probability of observing two or more events in a two-hour interval (6 P.M. to 8 P.M.). First, let's calculate the probability of observing 0 or 1 events in this time interval, and then we will find the complementary probability of two or more events occurring. For x = 0 events: \(P(X = 0) = \frac{e^{-\lambda(2)}(\lambda(2))^0}{0!} = \frac{e^{-4}(1)}{1} = e^{-4}\) For x = 1 event: \(P(X = 1) = \frac{e^{-\lambda(2)}(\lambda(2))^1}{1!} = e^{-4}(4) = 4e^{-4}\) Now we want to find the probability of two or more events occurring, which is the complement of the above probabilities: \(P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-4} - 4e^{-4} \approx 0.982\) Hence, the probability that two or more events occur between 6 P.M. and 8 P.M. is approximately 0.982.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability of Events
The probability of events in a Poisson process can seem daunting at first glance, but it's based on a simple premise: events occur independently and at a constant average rate. This rate is denoted as \( \lambda \) (lambda), representing the average number of events per time unit. For example, if \( \lambda = 2 \) events per hour, you'd typically expect two events to happen each hour on average.

Now, for a specified time period, we can calculate the likelihood of observing exactly \( x \) events using the Poisson probability formula:
\[ P(X = x) = \frac{e^{-\lambda t}(\lambda t)^x}{x!} \]
Where \( e \) is the base of natural logarithms, \( t \) is the length of the time interval in the same units as the rate \( \lambda \) and \( x! \) stands for \( x \) factorial. Once you understand the notation, calculating specific probabilities, like the likelihood of no events occurring within a time frame, becomes a straightforward substitution exercise.
Exponential Distribution
Delving deeper into the realm of statistics, the exponential distribution is closely tied to the concept of inter-arrival times in a Poisson process. It describes the time between successive events in a process that occurs continuously and independently at a constant average rate.

The exponential distribution has a memoryless property which means that the probability of an event occurring in the next interval is independent of how much time has already passed. The mean or expected value of the exponential distribution is given by \( E[T] = \frac{1}{\lambda} \) where \( \lambda \) is the rate of our Poisson process. This relationship allows us to predict the expected time for an event to occur, which is particularly useful for planning and reliability analysis.
Inter-Arrival Time
Inter-arrival time refers to the time elapsed between two successive events in a Poisson process. Understanding this concept helps in various fields such as telecommunications, traffic flow analysis, and customer service, where one needs to estimate the time until the next call, vehicle, or customer arrival. Since these time intervals follow an exponential distribution, we rely on the average rate \( \lambda \) to estimate the average inter-arrival time.

The beauty of the Poisson process lies in its simplicity: regardless of when the last event occurred, the expected time for the next event to happen remains consistently \( \frac{1}{\lambda} \) hours. Keep in mind, when calculating the expected time for multiple events, just sum up the expected times for each event.
Complementary Probability
Often while dealing with probabilities, it's easier to calculate the chance of something not happening and then deducing the opposite. This concept is known as complementary probability. It exploits the principle that the sum of the probabilities of all possible outcomes is 1.

For instance, when considering the probability of two or more events occurring within a certain timeframe in a Poisson process, it may be more straightforward to calculate the probability of the opposite outcomes (0 or 1 event) and subtract these from 1. This logical shortcut can simplify calculations and is a valuable part of a statistician's toolkit. By using \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \) we elegantly arrive at the probability for two or more events without directly calculating each individual scenario.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The water level of a certain reservoir is depleted at a constant rate of 1000 units daily. The reservoir is refilled by randomly occurring rainfalls. Rainfalls occur according to a Poisson process with rate \(.2\) per day. The amount of water added to the reservoir by a rainfall is 5000 units with probability \(.8\) or 8000 units with probability .2. The present water level is just slightly below 5000 units. (a) What is the probability the reservoir will be empty after five days? (b) What is the probability the reservoir will be empty sometime within the next ten days?

Customers can be served by any of three servers, where the service times of server \(i\) are exponentially distributed with rate \(\mu_{i}, i=1,2,3\). Whenever a server becomes free, the customer who has been waiting the longest begins service with that server. (a) If you arrive to find all three servers busy and no one waiting, find the expected time until you depart the system. (b) If you arrive to find all three servers busy and one person waiting, find the expected time until you depart the system.

An insurance company pays out claims on its life insurance policies in accordance with a Poisson process having rate \(\lambda=5\) per week. If the amount of money paid on each policy is exponentially distributed with mean \(\$ 2000\), what is the mean and variance of the amount of money paid by the insurance company in a four-week span?

The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?

A store opens at 8 A.M. From 8 until 10 customers arrive at a Poisson rate of four an hour. Between 10 and 12 they arrive at a Poisson rate of eight an hour. From 12 to 2 the arrival rate increases steadily from eight per hour at 12 to ten per hour at 2; and from 2 to 5 the arrival rate drops steadily from ten per hour at 2 to four per hour at \(5 .\) Determine the probability distribution of the number of customers that enter the store on a given day.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.