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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
Line probabilities after rework: \(A_1\): 36.2%, \(A_2\): 34.8%, \(A_3\): 29.0%.

Step by step solution

01

Define Events

Let the event \( A_i \) indicate that a component has been produced by line \( A_i \). Similarly, let \( R \) be the event that a component needs rework. We are tasked with finding the probabilities of \( A_1 \), \( A_2 \), and \( A_3 \) given that rework is needed, that is, \( P(A_1|R) \), \( P(A_2|R) \), and \( P(A_3|R) \).
02

Interpret Probability Data

From the problem description, we have the following probabilities: \( P(R|A_1) = 0.05 \), \( P(R|A_2) = 0.08 \), \( P(R|A_3) = 0.10 \). The overall percentages for each line are \( P(A_1) = 0.50 \), \( P(A_2) = 0.30 \), and \( P(A_3) = 0.20 \).
03

Compute Total Probability of Rework

Use 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 probabilities to find \( P(R) \):\[P(R) = (0.05)(0.50) + (0.08)(0.30) + (0.10)(0.20).\]Calculating, \( P(R) = 0.025 + 0.024 + 0.020 = 0.069 \).
04

Calculate Conditional Probabilities

Use Bayes' Theorem to find the desired probabilities:- \[ P(A_1|R) = \frac{P(R|A_1)P(A_1)}{P(R)} \]- \[ P(A_2|R) = \frac{P(R|A_2)P(A_2)}{P(R)} \]- \[ P(A_3|R) = \frac{P(R|A_3)P(A_3)}{P(R)} \]Substitute the values calculated:- \[ P(A_1|R) = \frac{0.025}{0.069} \approx 0.362 \]- \[ P(A_2|R) = \frac{0.024}{0.069} \approx 0.348 \]- \[ P(A_3|R) = \frac{0.020}{0.069} \approx 0.290 \].
05

Verify Solution

The condition \( P(A_1|R) + P(A_2|R) + P(A_3|R) \) should equal 1 to confirm that the conditional probabilities sum to 100%: \( 0.362 + 0.348 + 0.290 = 1 \). Thus, the calculations have been checked.

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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 way to calculate the probability of an event occurring, given that another event has already happened. In simpler terms, it answers the question: "How likely is event A to occur if event B is known to have happened?" This is denoted by the symbol \( P(A|B) \), which reads as "the probability of A given B".

In our exercise, we're interested in finding out the probabilities that a component needing rework came from a specific assembly line. Here, each assembly line occurrence (\( A_1, A_2, A_3 \)) is considered separately given the event \( R \) (rework is needed).
  • \( A_1 \): Component is from line A1.
  • \( A_2 \): Component is from line A2.
  • \( A_3 \): Component is from line A3.
  • \( R \): Component needs rework.
To calculate such probabilities, Bayes' Theorem is used, which adjusts our initial probabilities based on new information. In our scenario, this theorem helps establish what fraction of the defective pieces originates from each line.
Law of Total Probability
The law of total probability is a useful tool for computing the total probability of an event based on its association with multiple mutually exclusive scenarios or subsets. In simple terms, it breaks down a complicated situation into smaller, more manageable parts.

In our exercise, the total probability of a component needing rework \( P(R) \) is determined by considering each production line separately and then adding these probabilities together. Imagine breaking a task into smaller tasks: you add the probabilities of needing rework from each line weighed by how often each line is used.
  • \( P(R | A_1)P(A_1) \): Prob. of rework from line A1.
  • \( P(R | A_2)P(A_2) \): Prob. of rework from line A2.
  • \( P(R | A_3)P(A_3) \): Prob. of rework from line A3.
The sum of these computed probabilities gives us \( P(R) \), the overall likelihood of rework, totaling \( 0.069 \) or \( 6.9\% \) when expressed as a percentage.
Probability Calculations
Probability calculations involve using mathematical formulas to find the exact likelihood of event occurrences. These calculations can easily be handled with formulas and the precision of mathematics ensures accuracy of the results.

In our context, we applied probability calculations using the Bayes' Theorem to derive conditional probabilities that specify how likely a component needing rework came from each specific line. The formula used is:
- \( P(A_i|R) = \frac{P(R|A_i)P(A_i)}{P(R)} \)
Where:- \( P(A_i|R) \) is the probability that a component is from line \( A_i \) given it needs rework,- \( P(R|A_i) \) is the probability a component needs rework given it's from line \( A_i \),- \( P(A_i) \) is the probability a component is from line \( A_i \),- \( P(R) \) is the overall probability of rework.
After plugging in our values:
  • For line A1: \( P(A_1|R) = \frac{0.025}{0.069} \approx 0.362 \)
  • For line A2: \( P(A_2|R) = \frac{0.024}{0.069} \approx 0.348 \)
  • For line A3: \( P(A_3|R) = \frac{0.020}{0.069} \approx 0.290 \)
This illustrates a neat way how mathematics breaks down more complex reality into straightforward answers, detailing the source probabilities for components requiring rework.

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