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Chapter 2: Problem 3

A batch of 500 containers of frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement, from the batch. Let \(A\) and \(B\) denote the events that the first and second containers selected are defective, respectively. a. Are \(A\) and \(B\) independent events? b. If the sampling were done with replacement, would \(A\) and \(B\) be independent?

Short Answer

Expert verified
a) No, they are not independent. b) Yes, they would be independent.

Step by step solution

01

Determine Independence Under No Replacement

Two events are independent if the occurrence of one does not affect the probability of the other. We start with event \(A\): the probability of choosing a defective container first. There are 5 defective containers out of 500, so \(P(A) = \frac{5}{500} = \frac{1}{100}\). If \(A\) occurs, only 4 defective containers are left out of a total of 499 containers, making \(P(B|A) = \frac{4}{499}\). If \(A\) does not occur, then \(P(B|A^c) = \frac{5}{499}\). Hence, \(P(B)\) is not equal to \(P(B|A)\), which indicates \(A\) and \(B\) are not independent.
02

Determine Independence With Replacement

If sampling is done with replacement, the probability of each selection remains constant. In this case, \(P(A) = \frac{5}{500}\) and if a defective container is selected and replaced, \(P(B)\) also remains \(\frac{5}{500}\). The probability \(P(B|A)\) remains \(\frac{5}{500}\), showing \(P(B) = P(B|A)\). Thus, events \(A\) and \(B\) are independent when the sampling is done with replacement.

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Key Concepts

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

Understanding Probability
Probability measures how likely an event is to occur. It is expressed as a number between 0 and 1. If an event is certain to happen, its probability is 1. If it cannot happen, its probability is 0. The probability of an event, such as randomly selecting a defective container from a batch, is calculated by dividing the number of successful outcomes by the total number of possibilities.For example, in the case of selecting a defective container from a batch of 500 with only 5 defectives, the probability is:
  • Number of defective containers: 5
  • Total number of containers: 500
  • Probability of selecting a defective container: \( P(A) = \frac{5}{500} = \frac{1}{100} \)
Probability helps in understanding and predicting the likelihood of events in a given situation.
Event Independence
Event independence refers to how the occurrence of one event affects the probability of another event. Two events, say A and B, are said to be independent if the occurrence of A does not change the probability of B and vice versa. Mathematically, two events A and B are independent if:
  • \( P(B|A) = P(B) \) or
  • \( P(A|B) = P(A) \)
For example, if selecting one defective container does not change the chance of selecting another defective one, then events would be independent. In real-life scenarios, independence often helps in simplifying probability computations by allowing each event to be considered separately.
Sampling With Replacement
Sampling with replacement means that once an item is sampled, it is returned to the sample pool before the next draw. This implies that each draw is independent from the others as the conditions remain unchanged.For instance, consider selecting a defective container from a batch of 500 and then returning it before selecting again:
  • Probability of the first draw being defective is \( \frac{5}{500} \)
  • Return the item, and on the next draw, probability remains \( \frac{5}{500} \)
In this way, each sample does not affect the others, maintaining independence between events. This method is significant in probability because it simplifies calculations and keeps conditions consistent.
Sampling Without Replacement
Sampling without replacement implies that once an item is sampled, it is not returned to the sample pool. Each subsequent selection is affected by the previous selections. Suddenly, probabilities change as the number of items and composition of the pool alters with each draw:
  • If a defective container is selected first, probability changes for the next draw. \( P(B|A) = \frac{4}{499} \)
  • If not selected, the next probability remains different. \( P(B|A^c) = \frac{5}{499} \)
This method can significantly alter the likelihood of outcomes, especially in small sample spaces. Sampling without replacement often results in dependent events since each choice influences subsequent possibilities.

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Most popular questions from this chapter

A test of a printed circuit board uses a random test pattern with an array of 10 bits and each is equally likely to be 0 Or 1. Assume the bits are independent. a. What is the probability that all bits are \(1 \mathrm{~s}\) ? b. What is the probability that all bits are 0 s? c. What is the probability that exactly 5 bits are \(1 \mathrm{~s}\) and 5 bits are 0 s?

Suppose that \(P(A \mid B)=0.2, P\left(A \mid B^{\prime}\right)=0.3,\) and \(P(B)=0.8 .\) What is \(P(A) ?\)

A lot of 100 semiconductor chips contains 20 that are defective. a. Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective. b. Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

2.4 .8 A Web ad can be designed from four different colors, three font types, five font sizes, three images, and five text phrases. A specific design is randomly generated by the Web server when you visit the site. Let \(A\) denote the event that the design color is red. and let \(B\) denote the event that the font size is not the smallest one. Use the addition rules to calculate the following probabilities. a. \(P(A \cup B)\) b. \(P\left(A \cup B^{\prime}\right)\) c. \(P\left(A^{\prime} \cup B^{\prime}\right)\)

Semiconductor lasers used in optical storage products require higher power levels for write operations than for read operations. High-power-level operations lower the useful life of the laser. Lasers in products used for backup of higher-speed magnetic disks primarily write, and the probability that the useful life exceeds five years is \(0.95 .\) Lasers that are in products that are used for main storage spend approximately an equal amount of time reading and writing, and the probability that the useful life exceeds five years is \(0.995 .\) Now, \(25 \%\) of the products from a manufacturer are used for backup and \(75 \%\) of the products are used for main storage. Let \(A\) denote the event that a laser's useful life exceeds five years, and let \(B\) denote the event that a laser is in a product that is used for backup. Use a tree diagram to determine the following: a. \(P(B)\) b. \(P(A \mid B)\) c. \(P\left(A \mid B^{\prime}\right)\) d. \(P(A \cap B)\) e. \(P\left(A \cap B^{\prime}\right)\) f. \(P(A)\) g. What is the probability that the useful life of a laser exceeds five years? h. What is the probability that a laser that failed before five years came from a product used for backup?
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