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Chapter 5: Problem 29

A large retail store sells MP3 players. A customer who purchases an MP3 player can pay either by cash or credit card. An extended warranty is also available for purchase. Suppose that the events \(M=\) event that the customer paid by cash \(E=\) event that the customer purchased an extended warranty are independent with \(P(M)=0.47\) and \(P(E)=0.16\). a. Construct a "hypothetical 1000 " table with columns corresponding to cash or credit card and rows corresponding to whether or not an extended warranty is purchased. (Hint: See Example 5.9) b. Use the table to find \(P(M \cup E)\). Give a long-run relative frequency interpretation of this probability.

Short Answer

Expert verified
Using the probabilities given and after constructing the table, the probability that a customer pays in cash or purchases an extended warranty is computed to be 0.57. This means that in the long-run, out of every 1000 customers, about 570 customers are either likely to pay cash or buy the extended warranty or do both.

Step by step solution

01

Constructing the Table

Let's create a hypothetical 1000 purchases table, using the probabilities given: Cash payments (\(M\)) = 0.47 * 1000 = 470Credit card payments = 1000 - 470 = 530Extended warranty purchases (\(E\)) = 0.16 * 1000 = 160Without extended warranty = 1000 - 160 = 840The assumption of independence of events allows for calculation of joint occurrences in the following way:Cash and Warranty (\(M \cap E\)) = \(P(M) * P(E) * 1000 = 470 * 0.16 = 75.2 \approx 75\)Credit Card and Warranty = 1000 - 75 = 925Cash and No Warranty = 470 - 75 = 395 Credit Card without Warranty= 530 - 85= 445
02

Calculate the Probability

We've been asked to find the probability that a customer either pays by cash or buys the extended warranty (\(P(M \cup E)\)). In probability theory, this is found by: \(P(M \cup E) = P(M) + P(E) - P(M \cap E)\)Substituting gives us:= 0.47 + 0.16 - (0.47*0.16)= 0.57
03

Interpretation

This probability of 0.57 means that in the long-run, out of every 1000 customers, 570 of them are expected to either pay in cash or buy an extended warranty, or do both.

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Key Concepts

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

Independence of Events
In probability theory, independence of events refers to a situation where the occurrence of one event does not impact the occurrence of another. For example, in the context of a retail store, consider a customer鈥檚 choice of payment between cash and credit card, and whether they purchase an extended warranty. These choices are independent if buying an MP3 player with cash does not influence the decision of purchasing an extended warranty and vice versa.

Mathematically, this is expressed as:
  • Two events, say event A and event B, are independent if \( P(A \cap B) = P(A) \times P(B) \).
In our example, \( M \) represents paying by cash, and \( E \) represents purchasing an extended warranty. Given \( P(M) = 0.47 \) and \( P(E) = 0.16 \), they are independent because \( P(M \cap E) = 0.47 \times 0.16 \). This simplifies the analysis and calculations significantly, as knowledge about one event doesn't change the expectation of another.
Joint Probability
Joint probability involves finding the likelihood of two events happening at the same time. For instance, the probability of paying with cash and purchasing an extended warranty would be a joint probability. This is seen in the intersection of two events, mathematically described by \( P(A \cap B) \).

In our original exercise, the joint probability was calculated for cash payment and extended warranty as \( P(M \cap E) = 75.2 \approx 75 \). This means that out of 1000 hypothetical customers, around 75 are expected to pay by cash and also purchase an extended warranty. Understanding joint probability helps in figuring out how two independent events coincide or overlap with each other in a given scenario.
Union of Events
The union of events in probability theory is expressed as the probability that at least one of multiple possible events occurs. This is denoted by \( P(A \cup B) \), meaning event A happens or event B happens, or both occur. It鈥檚 a central concept when determining the probability of multiple scenarios happening together.

In the exercise provided, finding \( P(M \cup E) \) refers to calculating the likelihood that a customer either pays with cash, purchases an extended warranty, or does both. This is also known as the "or" probability. To calculate this, use the formula
  • \( P(M \cup E) = P(M) + P(E) - P(M \cap E) \)
Substituting the given probabilities, \( P(M \cup E) \) is found to be 0.57. Thus, in our "hypothetical 1000" table scenario, out of 1000 customers, we expect about 570 to either pay in cash, buy an extended warranty, or do both. Understanding the union of events is crucial for assessing aggregate probabilities.

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Most popular questions from this chapter

Suppose that \(20 \%\) of all teenage drivers in a certain county received a citation for a moving violation within the past year. Assume in addition that \(80 \%\) of those receiving such a citation attended traffic school so that the citation would not appear on their permanent driving record. Consider the chance experiment that consists of randomly selecting a teenage driver from this county. a. One of the percentages given in the problem specifies an unconditional probability, and the other percentage specifies a conditional probability. Which one is the conditional probability, and how can you tell? b. Suppose that two events \(E\) and \(F\) are defined as follows: \(E=\) selected driver attended traffic school \(F=\) selected driver received such a citation Use probability notation to translate the given information into two probability statements of the form \(P(\ldots)=\) probability value.

Consider a chance experiment that consists of selecting a student at random from a high school with 3,000 students. a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.

A Gallup survey found that \(46 \%\) of women and \(37 \%\) of men experience pain on a daily basis (San Luis Obispo Tribune, April 6,2000 ). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain on a daily basis independent or dependent? Explain.

If you were to roll a fair die 1,000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

Lyme disease is the leading tick-borne disease in the United States and Europe. Diagnosis of the disease is difficult and is aided by a test that detects particular antibodies in the blood. The article "Laboratory Considerations in the Diagnosis and Management of Lyme Borreliosis" (American Journal of Clinical Pathology [1993]: \(168-174\) ) used the following notation: + represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities were reported in the article:\(\begin{aligned} P(L) &=0.00207 \\ P\left(L^{C}\right) &=0.99793 \\ P(+\mid L) &=0.937 \\ P(-\mid L) &=0.063 \\ P\left(+\mid L^{C}\right) &=0.03 \\ P\left(-\mid L^{C}\right) &=0.97 \end{aligned}\) a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a "hypothetical \(1000 "\) table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use the table constructed in Part (b) to calculate this probability.
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