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Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a time line is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is 0.8 2\. If Maria completes her part on time, the probability that Alex completes on time is \(0.9,\) but if Maria is late, the probability that Alex completes on time is only 0.6 . 3\. If Alex completes his part on time, the probability that Juan completes on time is \(0.8,\) but if \(\mathrm{Alex}\) is late, the probability that Juan completes on time is only 0.5 . 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(0.9,\) but if Juan is late, the probability that Jacob completes on time is only 0.7 . Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time, and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

Short Answer

Expert verified
The short answer can't be defined in this case as each simulation would output a different probability. However, by following the proposed steps, the probability can be easily estimated.

Step by step solution

01

Understand the problem

In this problem, the probability of each student completing their task on time depends on the performance of the previous student. Maria's probability is independent while others depend on the person just before them. We need to simulate this scenario with at least 20 trials to estimate the probability that the project is completed on time.
02

Initialize the variable for successful trials

First, let's start by initializing a variable successful_trials to 0. This variable will keep track of the number of trials where all students complete the project on time.
03

Conduct the simulations

Run a loop for 20 trials. Inside each trial, generate a random number for each student's part. If Maria's number falls within 1-8, mark her as on time. Otherwise, she is late. This determines the chances for Alex. If Maria is on time and if Alex's number falls within 1-9, mark him as on time. If Maria is late, Alex's chances drop to 1-6. Repeat the process for Juan and Jacob. If all four students are on time, increment successful_trials by 1.
04

Calculate the estimated probability

After all trials have been conducted, divide the successful_trials by the total number of trials (20 in this case). This fraction will represent the estimated probability of the project being completed on time.

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in probability theory that describes the likelihood of an event occurring given that another event has already happened. In educational settings, it's a critical concept for students to understand, as it applies to a multitude of real-world situations. Take, for instance, our exercise, which involves a sequence of tasks that need to be completed on time by different students for a group project. The completion of each task depends on the completion of the prior one, making this a classic example of conditional probability.

When we say that Alex has a 0.9 probability of finishing on time if Maria does, but only a 0.6 if she's late, we're talking about conditional probability. The probability of Alex finishing on time changes based on the outcome of Maria's task. To express this in mathematical terms, we define the conditional probability of event B given event A as P(B|A) = P(A and B) / P(A), provided that P(A) is not zero. In the context of our exercise, P(A) could be the probability of Maria finishing on time, while P(B|A) would be the probability of Alex finishing on time given that Maria has finished on time.
Probability Estimation
Probability estimation involves determining the likelihood of an event through experiments or simulations, rather than theoretical calculations. This is particularly useful when dealing with complex scenarios where multiple factors are at play, making analytical solutions difficult. In our textbook problem, we use probability estimation via simulation to predict the chance the entire project is completed on time.

To estimate this probability, we conduct a series of trials—in our case, at least 20. During each trial, we use a random digit to simulate the completion of each student’s part of the project. By tallying the trials where all students finish on time and dividing by the total number of trials, we get an empirical estimate of the probability. This hands-on approach not only makes the learning experience more interactive but also ingrains the principles of probability into students' understanding through repetitive practice.
Probability Dependence
The idea of probability dependence is crucial when events are not isolated but rather influenced by the outcomes of previous events. In the provided exercise, the probability of each student completing their part on time depends on whether the preceding student did so. This interconnection is what we refer to as probability dependence.

In simple terms, if the probability of an event changes when another event occurs, the two events are dependent. For example, Juan's probability of being on time is 0.8 if Alex is on time and 0.5 if Alex is late. Therefore, Juan's timely completion is dependent on Alex's performance. This concept is central to understanding many processes in nature, technology, and even social dynamics where outcomes are interconnected. Our educational exercises aim to convey how to handle dependent probabilities in a straightforward and intuitive manner, often using simulations to illustrate these relationships with tangible examples.

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Most popular questions from this chapter

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E} .\) One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

The paper "Predictors of Complementary Therapy Use Among Asthma Patients: Results of a Primary Care Survey" (Health and Social Care in the Community [2008]: \(155-164)\) described a study in which each person in a large sample of asthma patients responded to two questions: Question 1: Do conventional asthma medications usually help your symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? Suppose that this sample is representative of asthma patients. Consider the following events: \(E=\) event that the patient uses complementary therapies \(F=\) event that the patient reports conventional medications usually help The data from the sample were used to estimate the following probabilities: $$P(E)=0.146 \quad P(F)=0.879 \quad P(E \cap F)=0.122$$ a. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(E\) and \(n o t E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. The probability that an asthma patient responds that conventional medications do not help and that patient uses complementary therapies. ii. The probability that an asthma patient responds that conventional medications do not help and that patient does not use complementary therapies. iii. The probability that an asthma patient responds that conventional medications usually help or the patient uses complementary therapies. c. Are the events \(E\) and \(F\) independent? Explain.

The article "Anxiety Increases for Airline Passengers After Plane Crash" (San Luis Obispo Tribune, November 13,2001 ) reported that air passengers have a 1 in 11 million chance of dying in an airplane crash. This probability was then interpreted as "You could fly every day for 26,000 years before your number was up." Comment on why this probability interpretation is misleading.

Automobiles that are more than 10 years old must pass a vehicle inspection to be registered in a particular state. The state reports the probability that a car more than 10 years old will fail the vehicle inspection is \(0.09 .\) Give a relative frequency interpretation of this probability.

A rental car company offers two options when a car is rented. A renter can choose to pre-purchase gas or not and can also choose to rent a GPS device or not. Suppose that the events \(A=\) event that gas is pre-purchased \(B=\) event that a GPS is rented are independent with \(P(A)=0.20\) and \(P(B)=0.15\). a. Construct a "hypothetical 1000 " table with columns corresponding to whether or not gas is pre-purchased and rows corresponding to whether or not a GPS is rented. b. Use the table to find \(P(A \cup B)\). Give a long-run relative frequency interpretation of this probability.

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