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Chapter 2: Problem 113

A batch of 25 injection-molded parts contains five that have suffered excessive shrinkage. (a) If two parts are selected at random, and without replacement, what is the probability that the second part selected is one with excessive shrinkage? (b) If three parts are selected at random, and without replacement, what is the probability that the third part selected is one with excessive shrinkage?

Short Answer

Expert verified
(a) \( \frac{1}{5} \); (b) \( \frac{14}{23} \)

Step by step solution

01

Identify Total and Defective Parts

The total number of parts is 25, and there are 5 parts with excessive shrinkage. Therefore, the non-defective parts total 20 (25 - 5 = 20).
02

Calculate Probability for Part (a)

For the first selection, the total number of parts is 25. Regardless of the first part's condition, there are 24 parts remaining. If the first part was not defective, there are still 5 defective parts remaining in 24. So the probability that the second part is defective is \( \frac{5}{24} \). If the first part was defective, there are 4 defective parts remaining, so the probability is \( \frac{4}{24} \). Since there's a total of 5 ways to get a defective part at first pick and 20 for a non-defective first pick, the total probability is calculated using weighted average: \( P = \frac{5}{25} \times \frac{4}{24} + \frac{20}{25} \times \frac{5}{24} \).
03

Perform Calculations for Part (a)

Evaluate the expression:\[ P = \frac{5}{25} \times \frac{4}{24} + \frac{20}{25} \times \frac{5}{24} = \frac{1}{5} \times \frac{1}{6} + \frac{4}{5} \times \frac{5}{24} = \frac{1}{30} + \frac{20}{120} = \frac{1}{30} + \frac{1}{6} = \frac{6}{180} + \frac{30}{180} = \frac{36}{180} = \frac{1}{5} \]
04

Calculate Probability for Part (b)

For the third selection, the probability depends on how many defective parts are picked first and second. There are several combinations, but only cases where the last pick is defective need tracking:- If two picked earlier are non-defective: \( \frac{20}{25} \times \frac{19}{24} \times \frac{5}{23} \)- If one picked earlier is defective, one is not: \( \frac{5}{25} \times \frac{20}{24} \times \frac{4}{23} + \frac{20}{25} \times \frac{5}{24} \times \frac{4}{23} \)Sum weighted probabilities.
05

Perform Calculations for Part (b)

Evaluate the expression:\[ P = \frac{20}{25} \times \frac{19}{24} \times \frac{5}{23} + 2 \times \frac{5}{25} \times \frac{20}{24} \times \frac{4}{23} = \frac{4}{5} \times \frac{19}{24} \times \frac{5}{23} + 2 \times \frac{1}{5} \times \frac{5}{6} \times \frac{4}{23} \]\[ = \frac{4 \times 19 \times 5}{5 \times 24 \times 23} + \frac{2 \times 4}{6 \times 23} \]\[ = \frac{76}{138} + \frac{8}{138} = \frac{84}{138} = \frac{42}{69} = \frac{14}{23} \]

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Key Concepts

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

Excessive Shrinkage
When we talk about excessive shrinkage in the context of manufacturing, it refers to parts that do not meet the expected size requirements due to inappropriate reduction in size during the production process. This is often a result of cooling at uneven rates in the manufacturing method, such as injection molding.

In our probability exercise, five out of twenty-five parts in a batch exhibit excessive shrinkage. Recognizing defective parts is crucial because it affects product quality and reliability. Quality assurance teams work diligently to identify and mitigate these flaws. Understanding the probability of selecting defective parts can help in planning and reducing waste.
Combination Probability
Combination probability focuses on finding the likelihood of a specific selection from a larger set. In our exercise, it helps determine the chances of picking parts with excessive shrinkage without replacement. Calculated using combinations, it helps us manage complex selections intuitively.

When calculating these probabilities, consider each selection event as connected to the ones before and after. As parts are picked and not replaced, the total number to choose from decreases, altering the probability ratios dynamically.
  • Probability changes dynamically as each item is picked.
  • No replacement means each selected part affects subsequent probabilities.
Understanding these changes is crucial for accurately determining outcomes in real-world scenarios.
Defective Parts Probability
Defective Parts Probability helps us understand the odds of selecting a faulty item from a batch. This is pivotal in quality control and inventory management. In our situation, we have a mixed batch of defective and non-defective parts.

To find the probability of a faulty part being selected, compute it step-by-step based on the sequence of selections. If you first select a non-defective item, the probability of picking a defective one next changes. Therefore, it鈥檚 crucial to address each step:-
  • Consider potential combinations of selections affecting defect chances.
  • Use sequential probabilities to refine defect risk understanding.
Calculated probabilities provide a predictive view of inventory worthiness and quality skillsets.
Weighted Average Probability
Weighted average probability combines results from different scenarios to find an overall probability. It's a method to reflect varying conditions in real-world situations, like our scenario with defective parts.

By weighing probabilities from different events by their likelihood of occurrence, we achieve a comprehensive look at overall chances. In our exercise, we used this concept to calculate the total chance the second or third part selected is defective.
  • Combine probabilities of different outcomes for a holistic opportunity understanding.
  • Crucial for planning and decision-making when probabilities vary across scenarios.
This means more informed decisions can be achieved in processes like manufacturing or quality assurance to ensure efficiency and effectiveness.

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