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Chapter 2: Problem 99

Fasteners used in aircraft manufacturing are slightly crimped so that they lock enough to avoid loosening during vibration. Suppose that \(95 \%\) of all fasteners pass an initial inspection. Of the \(5 \%\) that fail, \(20 \%\) are so seriously defective that they must be scrapped. The remaining fasteners are sent to a recrimping operation, where \(40 \%\) cannot be salvaged and are discarded. The other \(60 \%\) of these fasteners are corrected by the recrimping process and subsequently pass inspection. a. What is the probability that a randomly selected incoming fastener will pass inspection either initially or after recrimping? b. Given that a fastener passed inspection, what is the probability that it passed the initial inspection and did not need recrimping?

Short Answer

Expert verified
Part (a) 0.974; Part (b) approximately 0.9744.

Step by step solution

01

Calculate Initial Pass Probability

First, we determine the probability that a fastener passes the initial inspection. We are given that \(95\%\) of all fasteners pass initially. Therefore, \( P(\text{Initial Pass}) = 0.95 \).
02

Calculate Probability of Recrimping Pass

Calculate the chance for a fastener that fails initial inspection to eventually pass after recrimping.- \(5\%\) fail the initial inspection.- Of these, \(80\%\) are not scrapped (because \(20\%\) are scrapped).- \(60\%\) of those that go to recrimping succeed.So, the probability that a fastener goes through all these steps and passes after recrimping is \[ 0.05 \times 0.80 \times 0.60 = 0.024. \]
03

Overall Probability of Passing Inspection

To find the probability that a fastener passes inspection either initially or after recrimping, we sum the probabilities calculated in the first two steps: \[ P(\text{Pass}) = P(\text{Initial Pass}) + P(\text{Recrimping Pass}) = 0.95 + 0.024 = 0.974. \]
04

Calculate Conditional Probability

To answer part b, we find the probability that a fastener passed the initial inspection given that it passed inspection overall. Using the formula for conditional probability: \[P(A|B) = \frac{P(A \cap B)}{P(B)}.\] Here, \(A\) is passing the initial inspection and \(B\) is passing overall. Therefore:\[P(\text{Initial Pass | Pass}) = \frac{P(\text{Initial Pass})}{P(\text{Pass})} = \frac{0.95}{0.974} \approx 0.9744.\]
05

Conclusion: Answer for Both Parts

Based on the calculations, the probability that a fastener passed inspection either initially or after recrimping is \(0.974\). The probability that a fastener passed the initial inspection given that it passed the inspection overall is approximately \(0.9744\).

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Key Concepts

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

Conditional Probability
Understanding conditional probability is crucial for solving this exercise. Conditional probability helps us measure the likelihood of an event happening given that another event has already occurred.
For instance, you may want to know the probability of a fastener passing the initial inspection given that it passed the inspection overall. To find this, you use the formula:\[P(A|B) = \frac{P(A \cap B)}{P(B)},\] where \(A\) is the event of passing the initial inspection, and \(B\) is the event of passing the overall inspection.
By calculating this, you gain insights into dependencies between different stages in a process.
Probability Calculation
Probability calculation often involves several independent stages, as demonstrated in our exercise. Firstly, you want to determine the probability for an event such as passing the initial inspection. This is straightforward when the probability is given directly, like the \(95\%\) pass rate in the problem.
However, things get more complex when you deal with events dependent on others. For example, you compute the probability that a fastener requiring recrimping will eventually pass. You multiply the probabilities of all dependent events occurring together: initial failure, not being scrapped, and being successfully recrimped.
Multiply the probabilities for each condition accordingly: \[0.05 \times 0.80 \times 0.60 = 0.024.\]
Adding the probability of initially passing with that of passing after recrimping gives the overall pass probability: \[0.95 + 0.024 = 0.974.\]
Inspection Process
The inspection process is a series of checks ensuring the fasteners meet the required criteria. This process begins with an initial inspection, where \(95\%\) of the fasteners are deemed satisfactory. If a fastener fails, it is further examined to decide if it should be scrapped or undergo recrimping.
Understanding the inspection process involves assessing the likelihood of a fastener requiring additional work. While some may be defective beyond repair, most have a chance to be fixed through recrimping. This step-by-step assessment safeguards the product's final quality, maintaining industry standards.
Get familiar with these steps as they illustrate how quality assurance systems are rigorously managed and monitored.
Recrimping Operation
The recrimping operation is crucial in boosting product quality. If a fastener does not pass the initial inspection, it may undergo a recrimping process to correct potential deficiencies.
About \(20\%\) of the initially failing fasteners are scrapped due to severe defects. The rest enter the recrimping stage, where \(60\%\) of those can eventually pass through successful correction.
Recrimping ensures that even initially faulty products can meet safety and performance standards. Evaluating each step's success rate highlights the importance of reclamping in achieving acceptable pass rates for safety-critical components in industries like aircraft manufacturing.

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Most popular questions from this chapter

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities: a. \(P\) (all of the next three vehicles inspected pass) b. \(P\) (at least one of the next three inspected fails) c. \(P\) (exactly one of the next three inspected passes) d. \(P\) (at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

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The three most popular options on a certain type of new car are a built-in GPS \((A)\), a sunroof \((B)\), and an automatic transmission \((C)\). If \(40 \%\) of all purchasers request \(A, 55 \%\) request \(B, 70 \%\) request \(C, 63 \%\) request \(A\) or \(B, 77 \%\) request \(A\) or \(C, 80 \%\) request \(B\) or \(C\), and \(85 \%\) request \(A\) or \(B\) or \(C\), determine the probabilities of the following events. [Hint: " \(A\) or \(B\) " is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and not either of the other two options. d. The next purchaser will select exactly one of these three options.

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost \(\\}\). Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost, given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Mgmt., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]
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