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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

The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. The Price Waterhouse accounting firm conducted an independent audit by seeding the mail with letters and recording ontime delivery rates for these letters. Suppose that the results were as follows (these numhers are fictitions hut are compatible with summary values given in the article): $$ \begin{array}{lcc} & \begin{array}{l} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Lefters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \\ & & \\ \hline \end{array} $$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that two letters mailed in New York are both delivered on time d. The probability of on-time delivery nationwide

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