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Chapter 6: Problem 84

A company uses three different assembly lines \(-A_{1}\), \(A_{2}\), and \(A_{3}-\) to manufacture a particular component. Of those manufactured by \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components and \(10 \%\) of \(A_{3}\) 's components need rework. Suppose that \(50 \%\) of all components are produced by \(A_{1}\), whereas \(30 \%\) are produced by \(A_{2}\) and \(20 \%\) come from \(A_{3} .\) a. Construct a tree diagram with first-generation branches corresponding to the three lines. Leading from each branch, draw one branch for rework (R) and another for no rework (N). Then enter appropriate probabilities on the branches. b. What is the probability that a randomly selected component came from \(A_{1}\) and needed rework? c. What is the probability that a randomly selected component needed rework?

Short Answer

Expert verified
a. The tree diagram would have branches representing lines \(A_{1}\), \(A_{2}\), and \(A_{3}\) with additional branches representing 'Rework' and 'No Rework', each labeled with their corresponding probabilities. b. The probability that a randomly selected component came from \(A_{1}\) and needed rework is 0.025 or 2.5%. c. The probability that a randomly selected component needed rework is 0.071 or 7.1%.

Step by step solution

01

Construct a Tree Diagram

The first step is constructing a tree diagram to visualize the problem better. Each of the three assembly lines \(A_{1}\), \(A_{2}\), and \(A_{3}\) are the three initial branches. From each branch, two additional branches represent whether the component needs rework or not. The probabilities given in the problem are posted along the corresponding branches.
02

Calculate the Probability of a Component from \(A_{1}\) that Needs Rework

To find the probability that a component came from \(A_{1}\) and needed rework, we should multiply the probability that a component is made in \(A_{1}\) and the conditional probability that the component from \(A_{1}\) needed rework. This is given by: \(Pr(A_{1} and Rework) = Pr(A_{1}) * Pr(Rework|A_{1}) = 0.50 * 0.05 = 0.025, or 2.5%.
03

Calculate the Total Probability that a Component Needs Rework

To calculate the probability that a randomly selected component needs rework, we use the total probability theorem. Essentially, this sums up the probability of 'Rework' from all three assembly lines. This is given by: \(Pr(Rework) = Pr(A_{1}) * Pr(Rework|A_{1}) + Pr(A_{2}) * Pr(Rework|A_{2}) + Pr(A_{3}) * Pr(Rework|A_{3}) = 0.5 * 0.05 + 0.3 * 0.08 + 0.2 * 0.10 = 0.071, or 7.1%.

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Key Concepts

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

Total Probability Theorem
The total probability theorem is a fundamental rule in probability that allows us to determine the likelihood of an event by considering all possible pathways to that event. For instance, if we're trying to calculate the probability that a manufactured component will need rework, we'd consider the probability of it requiring rework when produced by each assembly line separately and then combine these probabilities.

Using the example from the exercise, with three assembly lines, we apply the theorem as follows:
  • Determine the probability that a component from each line needs rework.
  • Multiply this by the probability that a component comes from each line.
  • Sum these results to get the total probability of rework.
This approach ensures a comprehensive assessment of all production sources, leading to a more accurate measure of overall performance or risk.
Conditional Probability
Conditional probability deals with the likelihood of an event occurring given that another event has already happened. It's represented as \(Pr(B|A)\), which reads 'the probability of B given A.'

In the context of our manufacturing example, suppose we want to know the probability that a component needs rework given that it was produced by assembly line \(A_1\). We use the given data for the proportion of \(A_1\)'s output that needs rework (5%). The application of conditional probability allows us to focus on a subset within the whole system, providing crucial insights for targeted improvements and quality control. By analyzing conditional probabilities, manufacturers can identify which assembly lines might benefit from process optimization to reduce the need for rework.
Probability Tree Diagrams
Probability tree diagrams are visual tools used to map out the different outcomes of a sequence of events, making it easier to calculate combined probabilities. For our manufacturing problem, we would construct a tree with branches for each assembly line and further branches for 'rework' and 'no rework'.

By labeling each branch with the appropriate probabilities, we obtain a clear and organized presentation that guides us through the complexities of the system. Tree diagrams are particularly useful for illustrating how different probabilities interact, and they serve as visual aids to help with understanding and solving probability problems efficiently. They're an embodiment of the saying 'a picture is worth a thousand words,' especially in educational settings where visual learning can be more effective.
Statistics Education
The importance of statistics education cannot be overstated. It equips students with the tools to understand data, assess risks, and make informed decisions. The concepts of conditional probability and total probability theorem are core components of this education, as they are frequently used to discern patterns and predict outcomes in various fields.

By incorporating exercises like our assembly line problem into the curriculum, we prepare students for real-world situations where data-driven insights are crucial. Educators should strive to use clear explanations, step-by-step guides, and visual aids like tree diagrams to enhance comprehension. Ensuring that students grasp the foundational concepts in a statistics course sets them up for success in academic pursuits and professional endeavors where analysis and data interpretation are necessary skills.

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Most popular questions from this chapter

At a large university, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used a book by Professor Mean; during the winter quarter, 300 students used a book by Professor Median; and during the spring quarter, 200 students used a book by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with the text in the fall quarter, 150 in the winter quarter, and 160 in the spring quarter. a. If a student who took statistics during one of these three quarters is selected at random, what is the probability that the student was satisfied with the textbook? b. If a randomly selected student reports being satisfied with the book, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? (Hint: Use Bayes' rule to compute three probabilities.)

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 \- are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\) A beats \(B)=.7, P(\) A beats \(C)=.8\), \(P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that A wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that \(A\) wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than three licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, 9 are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as many of the requests as possible. For example, the city might fill requests for \(2,3,1\), and 3 licenses and then select a request for \(3 .\) Because there is only one license left, the last request selected would receive a license, but only one. a. An individual has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

The following case study was reported in the article "Parking Tickets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the \(U n\) known. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=1 / 44\). a. What reasoning did the expert use to arrive at the probability of \(1 / 44\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 44\) ? What effect does this error have on the probability of occurrence? Do you think that \(1 / 44\) is larger or smaller than the correct probability of occurrence?
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