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Ch. 4 - Discrete Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 4, Problem 4.2.41a

Manufacturing An assembly line produces 10,000 automobile parts. Twenty percent of the parts are defective. An inspector randomly selects 10 of the parts


a. Use the Multiplication Rule (discussed in Section 3.2) to find the probability that none of the selected parts are defective. (Note that the events are dependent.)

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Step 1: Understand the problem. We are tasked with finding the probability that none of the 10 randomly selected parts are defective, given that 20% of the 10,000 parts are defective. Since the events are dependent (selecting one part affects the probability of the next), we will use the Multiplication Rule for dependent events.
Step 2: Define the probabilities. The probability of selecting a non-defective part on the first draw is \( P(A_1) = \frac{8000}{10000} \), since 80% of the parts are non-defective. For the second draw, assuming the first part was non-defective, the probability becomes \( P(A_2 | A_1) = \frac{7999}{9999} \), and so on for subsequent draws.
Step 3: Apply the Multiplication Rule for dependent events. The rule states that the probability of all events occurring is the product of their conditional probabilities: \( P(A_1 \cap A_2 \cap \dots \cap A_{10}) = P(A_1) \cdot P(A_2 | A_1) \cdot P(A_3 | A_1 \cap A_2) \cdot \dots \cdot P(A_{10} | A_1 \cap A_2 \cap \dots \cap A_9) \).
Step 4: Write the specific probabilities for each draw. For the 10 draws, the probabilities are: \( P(A_1) = \frac{8000}{10000}, P(A_2 | A_1) = \frac{7999}{9999}, P(A_3 | A_1 \cap A_2) = \frac{7998}{9998}, \dots, P(A_{10} | A_1 \cap A_2 \cap \dots \cap A_9) = \frac{7991}{9991} \).
Step 5: Multiply the probabilities together. To find the overall probability that none of the selected parts are defective, calculate the product: \( P(A_1 \cap A_2 \cap \dots \cap A_{10}) = \frac{8000}{10000} \cdot \frac{7999}{9999} \cdot \frac{7998}{9998} \cdot \dots \cdot \frac{7991}{9991} \). This product gives the desired probability.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Multiplication Rule

The Multiplication Rule in probability is used to find the probability of the occurrence of multiple events. When events are dependent, the probability of the second event is affected by the outcome of the first. This rule states that the probability of two dependent events occurring is the product of the probability of the first event and the conditional probability of the second event given the first.
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Probability of Multiple Independent Events

Dependent Events

Dependent events are those where the outcome of one event affects the outcome of another. In this scenario, selecting a defective part changes the total number of parts available for subsequent selections, thus altering the probabilities. Understanding how these dependencies work is crucial for accurately calculating probabilities in situations like this.
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Probability of Multiple Independent Events

Probability of Defective Parts

In this context, the probability of selecting a defective part is 20%, meaning that out of 10,000 parts, 2,000 are expected to be defective. This probability is essential for calculating the likelihood of selecting non-defective parts. To find the probability that none of the selected parts are defective, one must consider the changing probabilities as parts are selected without replacement.
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Introduction to Probability
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