/*! 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 53 The water level of a certain res... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(0.2\) per day. The amount of water added to the reservoir by a rainfall is 5000 units with probability \(0.8\) or 8000 units with probability \(0.2 .\) 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?

Short Answer

Expert verified
Based on the above step-by-step solution, the short answer is: (a) The probability the reservoir will be empty after five days can be calculated as \(P(\text{Reservoir empty after 5 days}) = \sum_{k=0}^{5} P(E_k | X=k) P(X=k)\), where \(P(E_k | X=k)\) represents the probability of the reservoir being empty given \(k\) rainfalls and \(P(X=k)\) is the Poisson distribution probability for \(k\) rainfalls in 5 days. (b) The probability the reservoir will be empty sometime within the next ten days can be calculated as \(P(\text{Reservoir empty within 10 days}) = 1 - \prod_{t=1}^{10} (1-P(E_{X_t}))\), where \(P(E_{X_t})\) represents the probability of the reservoir being empty on the day \(t=1,2,\dots,10\).

Step by step solution

01

(Part (a): Number of Rainfalls)

First, we will find the probability of having \(k=0,1,..,5\) rainfalls within a 5-day period using the Poisson distribution formula: \(P(X=k) = \frac{(\lambda T)^k e^{-(\lambda T)}}{k!}\) where \(X\) represents the number of rainfalls, \(\lambda=0.2\) is the rate per day, \(T=5\) is the number of days, and \(k\) is the number of rainfalls we want to compute.
02

(Part (a): Water Level Condition)

Next, we will calculate the probability of the water level being empty with at least \(k=0,1,...,5\) rainfalls. The total depletion in 5 days is 5,000 units per day which is 25,000 units. The possible water refilled is 5,000 units with probability 0.8 and 8,000 units with probability 0.2. Let \(E_k\) denote the event of having the reservoir empty with \(k\) rainfalls. For each \(k\), we can compute the probability \(P(E_k)\) by considering all possible combinations of 5,000 units and 8,000 units amounts of water refilled. Afterward, we find the probability of the reservoir being empty as: \(P(\text{Reservoir empty after 5 days}) = \sum_{k=0}^{5} P(E_k | X=k) P(X=k)\)
03

(Part (b): Extend to 10 days)

Now, to calculate the probability of the reservoir being empty within the next 10 days (sometime in the first 10 days, not exactly on the tenth day), we will consider all possible scenarios. We will iterate through the days and compute the probability of the reservoir being empty on each day. For each day \(t=1,2,\dots,10\), we will first find the number of rainfalls until day \(t\), denoted by \(X_t\). Then, we will calculate the probability of the reservoir being empty on the day \(t\), \(P(E_{X_t})\), using the method in part (a). Finally, we will find the probability of the reservoir being empty sometime within the next 10 days as: \(P(\text{Reservoir empty within 10 days}) = 1 - P(\text{Reservoir not empty any day within 10 days}) = 1 - \prod_{t=1}^{10} (1-P(E_{X_t}))\)

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
Probability is a measure of how likely an event is to occur. In the context of a Poisson process, it's used to determine the likelihood of events happening over a fixed period. For example, if it rains according to a Poisson process, we can calculate the probability of different numbers of rainfall events occurring over a period, like five or ten days.

When considering whether the reservoir empties, we look at the probability of various scenarios of rainfall and water refill that would leave the reservoir empty. This involves both the chance of rainfalls occurring and how much water each rainfall adds. By understanding the probability, you can predict potential scenarios to manage water resources effectively. This is calculated using the Poisson formula:

  • \(P(X=k) = \frac{(\lambda T)^k e^{-(\lambda T)}}{k!}\)
Random Variables
Random variables are used to quantify random phenomena. In the given problem, the number of rainfalls is a discrete random variable. The random variable can take multiple values, such as 0, 1, 2, etc., representing the number of rainfalls within a specified timeframe.

Random variables help us map real-world random occurrences to mathematical concepts, allowing us to perform calculations and predictions. For example, the amount of water added to the reservoir from a single rainfall event—either 5000 or 8000 units—is also governed by a random variable that follows a certain probability distribution. Recognizing and using random variables are vital in handling such stochastic processes efficiently.
Expected Value
The expected value is a key concept describing the average outcome of a random process over a large number of trials. In the case of rainfalls, it helps predict the average water contribution from rain over a period.

The expected value can be calculated for complex situations, like the mix of rainfalls with different water contributions. For instance, if each rainfall contributes 5000 units with a probability of 0.8, and 8000 units with a probability of 0.2, the expected amount from one rainfall is:

\[E = (5000 \times 0.8) + (8000 \times 0.2) = 5600 \text{ units}\]

This expected value helps determine how much water can be anticipated overall, aiding in decision-making about reservoir management.
Event Occurrence
Event occurrence pertains to understanding when and how frequently certain events happen, such as rainfalls that color the water management in the reservoir problem. The Poisson process is central here, providing a way to model events that are random but occur with a known average rate.

This model suits scenarios like rainfall, where events happen independently and unpredictably, but with some regularity over time. By examining event occurrence, planners can estimate not only the likelihood of the reservoir running dry, as in part (a) and (b) of the exercise, but also other potential outcomes—like overflows or optimal refill rates, which help better prepare for fluctuating water levels.

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

Policyholders of a certain insurance company have accidents at times distributed according to a Poisson process with rate \(\lambda .\) The amount of time from when the accident occurs until a claim is made has distribution \(G\). (a) Find the probability there are exactly \(n\) incurred but as yet unreported claims at time \(t\). (b) Suppose that each claim amount has distribution \(F\), and that the claim amount is independent of the time that it takes to report the claim. Find the expected value of the sum of all incurred but as yet unreported claims at time \(t\).

Consider a conditional Poisson process in which the rate \(L\) is, as in Example \(5.29\), gamma distributed with parameters \(m\) and \(p\). Find the conditional density function of \(L\) given that \(N(t)=n\).

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?

Consider a nonhomogeneous Poisson process whose intensity function \(\lambda(t)\) is bounded and continuous. Show that such a process is equivalent to a process of counted events from a (homogeneous) Poisson process having rate \(\lambda\), where an event at time \(t\) is counted (independent of the past) with probability \(\lambda(t) / \lambda ;\) and where \(\lambda\) is chosen so that \(\lambda(s)<\lambda\) for all \(s\).

Customers arrive at a bank at a Poisson rate \(\lambda .\) Suppose two customers arrived during the first hour. What is the probability that (a) both arrived during the first 20 minutes? (b) at least one arrived during the first 20 minutes?

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.