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An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. a. Use the given probability information to set up a hypothetical 1000 table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. ii. the probability that a randomly selected customer purchases an extended warranty for neither the washer nor the dryer.

Short Answer

Expert verified
In summary, the probability of a customer purchasing extended warranties for both the washer and the dryer is 0.40, while the probability of a customer purchasing neither extended warranty is 0.41.

Step by step solution

01

Given probabilities

We are given that: - 52% of customers buy the extended warranty for the washer. - 47% of customers buy the extended warranty for the dryer. - 59% of customers buy at least one of the two extended warranties. Let W represent the event of buying a washer warranty, D represent the event of buying a dryer warranty.
02

Calculate probabilities of joint events

We know that the probability of either event W or event D occurring is given by: \(P(W \cup D) = P(W) + P(D) - P(W \cap D)\) We are given: \(P(W \cup D) = 0.59\) \(P(W) = 0.52\) \(P(D) = 0.47\) Solving for \(P(W \cap D)\), we get: \(P(W \cap D) = P(W) + P(D) - P(W \cup D) = 0.52 + 0.47 - 0.59 = 0.40\)
03

Constructing the hypothetical 1000 table

We will now create a hypothetical 1000 table, assuming there are 1000 customers: | | Washer Warranty | No Washer Warranty | Total | |-----------------|------------------|--------------------|-------| | Dryer Warranty | 400 | 70 | 470 | | No Dryer Warranty | 120 | 410 | 530 | | Total | 520 | 480 | 1000 | b. Calculating the requested probabilities.
04

Probability of purchasing both warranties

To find the probability that a randomly selected customer purchases both washer and dryer extended warranties, look at the value in the table corresponding to both events happening: 400. Now, divide this value by the total number of customers (1000) to get the probability: \(P(W \cap D) = \frac{400}{1000} = 0.40\) (which we already calculated earlier)
05

Probability of purchasing neither warranty

To find the probability that a randomly selected customer purchases neither washer nor dryer extended warranties, look at the value in the table corresponding to both events not happening: 410. Now, divide this value by the total number of customers (1000) to get the probability: \(P(\overline{W} \cap \overline{D}) = \frac{410}{1000} = 0.41\) So, i. The probability of a customer purchasing both warranties is 0.40. ii. The probability of a customer purchasing neither warranty is 0.41.

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Key Concepts

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

Extended Warranties
When making a purchase, consumers are often offered extended warranties, which provide additional protection and extend the life of the warranty beyond the standard period. In the context of probability calculations, statistics about extended warranties can illustrate important concepts such as the likelihood of consumers opting for such extra coverage. Extended warranties can be viewed as events in a sample space where we analyze how many customers choose more security for their products. This ties closely to understanding customer behavior and can help businesses develop better marketing strategies for these add-on services.

In our exercise, extended warranties for washers and dryers serve as focal events for which we calculate the probability of customers purchasing them. It is crucial to communicate these concepts in a manner that highlights their practical implications, reinforcing how the subject matter applies in real-life situations.
Hypothetical 1000 Table
A hypothetical 1000 table is a tool used to visualize and simplify complex probability problems by imagining a sample of 1000 instances, which makes calculating probabilities more intuitive. This approach is particularly helpful when dealing with percentages, as it translates them into whole numbers that are easier to comprehend.

In our scenario, we create a hypothetical table based on the given probabilities. The rows and columns correspond to whether a customer has bought an extended warranty for the washer, the dryer, both, or neither. The table helps us see the distribution of 1000 customers' choices regarding extended warranties and calculate joint and exclusive probabilities from it. Clear communication regarding the setting up and the utility of such a table can foster better understanding of how to envision and solve probability-related problems.
Joint Probability
Joint probability is a measure of two events occurring simultaneously. In our exercise, this corresponds to the probability of customers purchasing both a washer and a dryer extended warranty. To calculate the joint probability, we use the formula:
\[P(W \cap D) = P(W) + P(D) - P(W \cup D)\]
Here, \(P(W \cap D)\) is the joint probability of events W (washer warranty) and D (dryer warranty) occurring together. Understanding how to manipulate this formula is critical in deriving accurate probabilities and making informed decisions based on those probabilities.

How Joint Probability is Calculated

Using given percentages, we subtract the total probability of purchasing at least one warranty from the sum of the probabilities of purchasing each warranty individually. This calculation provides us with the joint probability, streamlined with our hypothetical 1000 table for ease of understanding.
Mutually Exclusive Events
Mutually exclusive events are events that cannot occur at the same time. For example, flipping a coin cannot result in both a head and a tail simultaneously. Understanding mutually exclusive events is essential in calculating probabilities, as it affects how we add probabilities of individual events to find the total probability of any event occurring.

In the case of the appliance manufacturer's extended warranties, purchasing a warranty for the washer does not preclude a customer from also purchasing one for the dryer. Therefore, these are not mutually exclusive events. Clarifying this concept is vital since the miscalculation of event exclusivity can lead to incorrect probability assessments. Emphasizing the difference between mutually exclusive and non-mutually exclusive events helps students avoid common mistakes when analyzing probabilities in a given scenario.

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Most popular questions from this chapter

An online store offers two methods of shipping-regular ground service and an expedited 2 -day shipping. Customers may also choose whether or not to have the purchase gift wrapped. Suppose that the events \(E=\) event that the customer chooses expedited shipping \(G=\) event that the customer chooses gift wrap are independent with \(P(E)=0.26\) and \(P(G)=0.12\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not expedited shipping was chosen and rows corresponding to whether or not gift wrap was selected. b. Use the information in the table to calculate \(P(E \cup G)\). Give a long- run relative frequency interpretation of this probability.

To help ensure the safety of school classrooms, the local fire marshal does an inspection at Thomas Jefferson High School each month to check for faulty wiring, overloaded circuits, and other fire code violations. Each month, one room is selected for inspection. Suppose that the probability that the selected room is a science classroom (biology, chemistry, or physics) is 0.6 and the probability that the selected room is a chemistry room is 0.4 . Use probability formulas to find the following probabilities. a. The probability that the selected room is not a science room. b. The probability that the selected room is a chemistry room and a science room. c. The probability that the selected room is a chemistry room given that the room selected was a science room. d. The probability that the selected room was a chemistry room or a science room.

5.62 An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. In Exercise \(5.34,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer.

An electronics store sells two different brands of DVD players. The store reports that \(30 \%\) of customers purchasing a DVD choose Brand \(1 .\) Of those that choose Brand \(1,20 \%\) purchase an extended warranty. Consider the chance experiment of randomly selecting a customer who purchased a DVD player at this store. 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 \(B\) and \(E\) are defined as follows: \(B=\) selected customer purchased Brand 1 \(E=\) selected customer purchased an extended warranty Use probability notation to translate the given information into two probability statements of the form \(P(\longrightarrow)=\) probability value.

The report "Improving Undergraduate Learning" (Social Science Research Council, 2011) summarizes data from a survey of several thousand college students. These students were thought to be representative of the population of all U.S. college students. When asked about a typical semester, \(68 \%\) said they would be taking a class that is reading intensive (requires more than 40 pages of reading per week). Only \(50 \%\) said they would be taking a class that is writing intensive (requires more than 20 pages of writing over the course of the semester). The percentage who said that they would be taking both a reading intensive course and a writing intensive course in a typical semester was \(42 \% .\) In Exercise \(5.40,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that a randomly selected student would be taking at least one reading intensive or writing intensive course. b. The probability that a randomly selected student would be taking a reading intensive course or a writing intensive course, but not both. c. The probability that a randomly selected student is taking neither a reading intensive nor a writing intensive course.

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