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Chapter 2: Problem 6

A batch of 500 containers for frozen orange juice contains 5 that are defective. Three are selected, at random, without replacement from the batch. a. What is the probability that the second one selected is defective given that the first one was defective? b. What is the probability that the first two selected are defective? c. What is the probability that the first two selected are both acceptable? d. What is the probability that the third one selected is defective given that the first and second ones selected were defective? e. What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? f. What is the probability that all three selected ones are defective?

Short Answer

Expert verified
1. \( \frac{4}{499} \); 2. \( \frac{20}{249500} \); 3. \( \frac{495}{500} \times \frac{494}{499} \); 4. \( \frac{3}{498} \); 5. \( \frac{4}{498} \); 6. \( \frac{60}{124251000} \).

Step by step solution

01

Understanding the problem

We are dealing with a probability problem involving 500 containers where 5 are defective. We will calculate probabilities for different scenarios of selecting defective and non-defective containers without replacement.
02

Probability that the second one is defective given the first is defective

If the first container is defective, then there are 4 defective containers left out of 499 total. The probability of selecting a defective container second is \( \frac{4}{499} \).
03

Probability that the first two are defective

For the first selection, the probability is \( \frac{5}{500} \), and for the second is \( \frac{4}{499} \), giving a combined probability of \( \frac{5}{500} \times \frac{4}{499} = \frac{20}{249500} \).
04

Probability that the first two are both acceptable

For the first selection, the probability of selecting a non-defective container is \( \frac{495}{500} \), and for the second \( \frac{494}{499} \), giving \( \frac{495}{500} \times \frac{494}{499} \).
05

Probability that the third one is defective given first two were defective

If the first two are defective, there are 3 defective containers left out of 498. The probability is \( \frac{3}{498} \).
06

Probability that the third one is defective given the first defective and second non-defective

After drawing one defective and one acceptable, we have \( \frac{4}{498} \) chance that the third container is defective.
07

Probability that all three selected are defective

Multiply the probabilities: \( \frac{5}{500} \times \frac{4}{499} \times \frac{3}{498} \). The calculation is \( \frac{60}{124251000} \).

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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 key concept in probability theory, used to determine the probability of an event occurring given that another event has already occurred. In our problem, questions like "What is the probability that the second container is defective given that the first one was defective?" involve conditional probability. To solve this, we examine how the occurrence of the first event affects the second. For instance:
  • When the first container selected is defective, the number of remaining defective containers decreases, altering the probability for the subsequent selections.
  • The probability that the second one is defective, given the first is defective, is calculated by considering that only 4 defective containers remain out of 499.
  • This calculation is \( \frac{4}{499} \) for the second selection being defective given the first was defective.
This method narrows down the pool of choices and simplifies probability calculations.
Combinatorics
Combinatorics plays a significant role in determining how many ways events can occur, particularly in problems involving selections like the one presented in the exercise. By using combinatorial principles, we can figure out the total number of possible combinations when selecting containers. In the context of our exercise:
  • The total number of combinations for any selection from 500 containers can be determined using combinations formulas, but here it's much simpler since we are selecting consecutively without formule.
  • Understanding that each selection without replacement changes the total pool of possible choices significantly helps interpret subsequent probabilities.
  • For example, selecting three containers one after another represents a sequence of dependent events, where each selection influences the next.
Using this orderly approach simplifies the calculations and understanding of the probabilities associated with each event.
Probability Theory
Probability theory is the mathematical foundation for quantifying uncertainty and making predictions about the occurrence of various events. This exercise utilizes probability theory to work through scenarios of selecting defective and non-defective containers. Let's look at the basics that can enhance understanding:
  • Probability is fundamentally calculated as the number of favorable outcomes divided by the total number of possible outcomes.
  • In the context of this problem, consider each container's probability as being defective or non-defective, depending on prior selections.
  • Through multiplication of probabilities, such as finding the joint probability of two or more events, we determine the likelihood of multiple events occurring in succession.
By applying these principles, students can methodically tackle even more complex probability challenges with confidence.
Defective Items Analysis
Defective items analysis involves a focused examination of defects within a sample or batch, often crucial in quality control and logistics. This exercise calls for understanding how defects influence probability calculations when selecting items from a batch. To dive deeper:
  • Recognize that knowing the proportion of defective items helps us evaluate the risk or likelihood of picking such an item. Here, with 5 out of 500 containers being defective, this is the base rate of defects.
  • When items are chosen without replacement, each draw alters the odds, as the total number of items and the fraction of defective ones shifts.
  • The sequential aspect of the selections highlights how each choice impacts subsequent probabilities, which is central to effectively analyzing defective items.
This type of analysis is particularly useful for businesses looking to minimize losses due to defects and optimize their selection processes.

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Most popular questions from this chapter

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