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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 line \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components need rework and \(10 \%\) of \(A_{3}\) 's need rework. Suppose that \(50 \%\) of all components are produced by line \(A_{1}\), \(30 \%\) are produced by line \(A_{2}\), and \(20 \%\) come from line \(A_{3}\). If a randomly selected component needs rework, what is the probability that it came from line \(A_{1}\) ? From line \(A_{2}\) ? From line \(A_{3}\) ?

Short Answer

Expert verified
The probabilities are \(0.3623\), \(0.3478\), and \(0.2899\) for lines \(A_1\), \(A_2\), and \(A_3\) respectively.

Step by step solution

01

Identifying Given Probabilities

Let Event \( A_i \) denote that the component is from assembly line \( A_i \). Then, the probabilities are: \( P(A_1) = 0.5 \), \( P(A_2) = 0.3 \), \( P(A_3) = 0.2 \). Let Event \( R \) denote that the component needs rework. The conditional probabilities are: \( P(R|A_1) = 0.05 \), \( P(R|A_2) = 0.08 \), \( P(R|A_3) = 0.10 \).
02

Calculating Total Rework Probability

We need to find the total probability that a component needs rework, \( P(R) \). This is given by the law of total probability:\[ P(R) = P(R|A_1)P(A_1) + P(R|A_2)P(A_2) + P(R|A_3)P(A_3).\]Substitute the given values:\[ P(R) = (0.05)(0.5) + (0.08)(0.3) + (0.10)(0.2) = 0.025 + 0.024 + 0.02 = 0.069. \]
03

Using Bayes' Theorem for Line A1

We want to find \( P(A_1|R) \), which is the probability that a component came from line \( A_1 \) given that it needs rework. Using Bayes' Theorem:\[ P(A_1|R) = \frac{P(R|A_1)P(A_1)}{P(R)}. \]Substitute the values:\[ P(A_1|R) = \frac{0.025}{0.069} \approx 0.3623. \]
04

Using Bayes' Theorem for Line A2

Next, use Bayes' Theorem to find \( P(A_2|R) \):\[ P(A_2|R) = \frac{P(R|A_2)P(A_2)}{P(R)}. \]Substitute the values:\[ P(A_2|R) = \frac{0.024}{0.069} \approx 0.3478. \]
05

Using Bayes' Theorem for Line A3

Finally, use Bayes' Theorem to find \( P(A_3|R) \):\[ P(A_3|R) = \frac{P(R|A_3)P(A_3)}{P(R)}. \]Substitute the values:\[ P(A_3|R) = \frac{0.02}{0.069} \approx 0.2899. \]

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Key Concepts

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

Law of Total Probability
The Law of Total Probability is an essential concept in probability theory that helps us calculate the probability of an event by considering all possible scenarios or paths that might lead to it. In simpler terms, it breaks down a problem into more accessible parts by considering the sum of probabilities of different ways an event can occur.

In the context of our exercise, we first identify the different assembly lines (A_1, A_2, and A_3) and their respective contributions to the total production. Each line has a different probability of producing a component that needs rework, given by the probabilities \(P(R|A_1)\), \(P(R|A_2)\), and \(P(R|A_3)\).

The law tells us the total probability that a component needs rework, \(P(R)\), can be calculated by considering the probability of needing rework from each line, weighted by the probability of a component coming from that line. This means we multiply each line’s probability of producing a faulty piece by the proportion of components coming from that line, and add them all up. In this exercise, we find that the total probability a component needs rework is \(P(R) = 0.069\). This total probability is crucial, as it sets the stage for using Bayes' theorem later on.
Conditional Probability
Conditional Probability deals with the likelihood of an event or outcome occurring, assuming another event has already taken place. It bridges the gap between independent events and dependent outcomes, helping us understand probabilities in complex scenarios.

To comprehend this fully, think about our probability problem. We're dealing with parts that may need rework, and we want to figure out which line a part likely comes from if it's faulty. We use conditional probability for this analysis. The exercise gives us initial probabilities such as \(P(R|A_1) = 0.05\); this means that there's a 5% chance a part from line \(A_1\) needs rework. Similarly, if it's from line \(A_2\), there's an 8% chance, and for \(A_3\), a 10% chance.

Conditional probability equips us with a formula that adjusts probabilities based on newly-acquired information. Rather than treating potential rework occurrences as isolated events, we integrate what we know about production lines to make informed predictions about the faulty component's origin. This forms the basis for applying Bayes' Theorem, as it involves updating an initial probability estimate based on new evidence (like the occurrence of defects).
Probability Theory
Probability Theory is the mathematical foundation that enables us to measure and interpret the likelihood of various outcomes. As a broader field, it includes numerous principles and rules, such as the Law of Total Probability and Conditional Probability, which are building blocks for more complex applications.

In probability theory, we often deal with random variables and events, using them to model real-world problems. The scenario with the manufacturing lines can be dissected quite effectively with these principles. We start with known probabilities for rework from each line and the fractions of total components coming from each line. These are our tools.

By correctly applying probability theory, we manage to explore how different independent events interact. Bayes' Theorem, employed in the original exercise, is a revolutionary part of this field. It specifically allows us to update our beliefs about where a component likely comes from when we learn it needs rework. This process of inference and adjustment is at the heart of probability theory, ensuring we can make informed predictions and decisions based on incomplete information.

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Most popular questions from this chapter

At a large university, in the never-ending quest for a satisfactory textbook, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used the text by Professor Mean; during the winter quarter, 300 students used the text by Professor Median; and during the spring quarter, 200 students used the text by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with Mean's book, 150 were satisfied with Median's book, and 160 were satisfied with Mode's book. If a student who took statistics during one of these quarters is selected at random and admits to having been satisfied with the text, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? [Hint: Draw a tree- diagram or use Bayes' theorem.]

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