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Chapter 2: Problem 77

An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. a. If \(20 \%\) of all seams need reworking, what is the probability that a rivet is defective? b. How small should the probability of a defective rivet be to ensure that only \(10 \%\) of all seams need reworking?

Short Answer

Expert verified
(a) The probability is 0.005935. (b) It should be 0.003443.

Step by step solution

01

Define the Problem

We need to find the probability that a single rivet is defective, given the probability that a seam needs reworking. We deal with 25 rivets per seam, and if any of these is defective, the seam needs reworking. Therefore, we need to calculate the probability that at least one rivet is defective per seam.
02

Use Complement Rule for (a)

Let's first deal with part (a). Let \( p \) be the probability that a single rivet is defective. The probability that at least one rivet in the seam of 25 is defective is \( 0.2 \). Using the complement rule, the probability that all 25 rivets are non-defective is \( 1 - 0.2 = 0.8 \). So, we have the equation: \( (1 - p)^{25} = 0.8 \).
03

Solve for Probability of Defective Rivet in (a)

To find \( p \), we rearrange the equation \( (1 - p)^{25} = 0.8 \) as follows:\[ 1 - p = 0.8^{1/25} \]Taking the 25th root of 0.8:\[ 1 - p = 0.994065 \]Thus:\[ p = 1 - 0.994065 = 0.005935 \]
04

Use Complement Rule for (b)

For part (b), we need to ensure that only \(10\%\) of all seams need reworking. Therefore, we want \( (1 - p)^{25} = 0.9 \).
05

Solve for Smaller Probability in (b)

Rearrange the equation \( (1 - p)^{25} = 0.9 \) as:\[ 1 - p = 0.9^{1/25} \]Taking the 25th root of 0.9:\[ 1 - p = 0.996557 \]Thus:\[ p = 1 - 0.996557 = 0.003443 \]

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Key Concepts

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

Defective Products
When discussing defective products in engineering, particularly in scenarios where a large batch of identical items is produced, it's crucial to understand the impact of defects on the overall product reliability. In our case, we are looking at rivets used in aircraft seams. A defective rivet could compromise the integrity of the entire seam, necessitating rework. Thus, even a single defective rivet among a set of 25 could require the seam to be redone.

Understanding how defects affect systems helps engineers implement quality control measures. This involves statistical models that help predict defects based on the probability of individual failures. When you deal with defective products, it is not only about identifying the probability of a defect but also understanding how it affects the larger system, such as a seam in an aircraft. Thus, identifying and minimizing the probability of defects is essential in maintaining high-quality standards in engineering.
Complement Rule
The Complement Rule is a fundamental concept in probability, used to simplify the calculation of probabilities of complex events. This rule states that the probability of an event happening is 1 minus the probability of it not happening. For example, in our scenario with rivets, if we know that a seam doesn't need reworking 80% of the time, then it must need reworking 20% of the time.

In mathematical terms, if you define an event where a seam doesn鈥檛 have any defects as non-defective, represented by the probability of all rivets being non-defective, then we have:
  • Probability of a non-defective event = 0.8
  • Probability of a defective event (at least one defective rivet) = 1 - Probability of non-defective event = 0.2
The Complement Rule becomes particularly powerful in scenarios like these, with large sample sizes, as it allows determining the probability of complex events through their complements, which are often easier to calculate.
Statistical Problem Solving
In the realm of probability in engineering, statistical problem solving is the key to predicting and improving product reliability. This involves using mathematical models and statistical tools to tackle problems related to defects and quality assurance. Break down a problem into manageable parts. For example, by identifying the probability of a single rivet being defective, we can then calculate the probability of needing a seam rework.

Engineers use statistical techniques like probability distributions and the Complement Rule to draw conclusions about the reliability of components. For rivets, we use these concepts to calculate how changing the probability of a single rivet being defective influences the likelihood of reworking seams. By solving equations and using iterative methods, engineers can optimize processes to ensure that quality standards are consistently met, thereby minimizing the need for extensive reworks and ensuring the reliability of final products.

These techniques ultimately help in understanding the underlying patterns and trends in defect occurrence, empowering engineers to enhance quality and prevent defects before they occur.

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Most popular questions from this chapter

Suppose an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over \(6 \mathrm{ft}\) in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A)\) ? Why?

A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let \(p\) denote the probability that the flaw is detected during any one fixation (this model is discussed in "Human Performance in Sampling Inspection," Human Factors, 1979: 99-105). a. Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? b. Give an expression for the probability that a flaw will be detected by the end of the \(n\)th fixation. c. If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection? d. Suppose \(10 \%\) of all items contain a flaw \([P\) (randomly chosen item is flawed) \(=.1]\). With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)? e. Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for \(p=.5\).

Use Venn diagrams to verify the following two relationships for any events \(A\) and \(B\) (these are called De Morgan's laws): a. \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\) b. \((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

The probability that a grader will make a marking error on any particular question of a multiple-choice exam is .1. If there are ten questions and questions are marked independently, what is the probability that no errors are made? That at least one error is made? If there are \(n\) questions and the probability of a marking error is \(p\) rather than \(.1\), give expressions for these two probabilities.
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