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Chapter 5: Problem 62

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.

Short Answer

Expert verified
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 is \(0.40\). b) The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer is \(0.41\).

Step by step solution

01

Find P(A 鈭� B) using the formula for the probability of the union of two events

Using the formula for the probability of the union of two events, we have: P(A 鈭� B) = P(A) + P(B) - P(A 鈭� B) Now, we will re-arrange the formula to find P(A 鈭� B): P(A 鈭� B) = P(A) + P(B) - P(A 鈭� B) Plug in the values given in the problem into the formulas: P(A 鈭� B) = 0.52 + 0.47 - 0.59 Calculate P(A 鈭� B): P(A 鈭� B) = 0.40 So, the probability that a randomly selected customer buys an extended warranty for both the washer and the dryer is 0.40.
02

Find P(A' 鈭� B') using the complement rule

We will use the complement rule to find P(A' 鈭� B'): P(A' 鈭� B') = 1 - P(A 鈭� B) Plug in the value of P(A 鈭� B) given in the problem into the formula: P(A' 鈭� B') = 1 - 0.59 Calculate P(A' 鈭� B'): P(A' 鈭� B') = 0.41 So, the probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer is 0.41. In conclusion, we have found the following 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: 0.40. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer: 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 you buy appliances like washers and dryers, you might be offered something called an "extended warranty". This is a type of service agreement that goes beyond the standard warranty offered by the manufacturer. It usually covers repairs or replacements after the initial warranty period ends.

The purpose of these warranties is to provide peace of mind. If anything goes wrong with your appliance beyond the initial warranty period, the extended warranty can help cover repair costs.
  • It usually starts when the manufacturer's warranty ends.
  • Covers various components and sometimes labor costs for repairs.
  • Can add one to several years of coverage to your appliance purchase.
In the context of our exercise, understanding who opts for these warranties helps calculate certain probabilities related to purchases.
Complement Rule
The complement rule in probability is a handy tool that helps us find the probability of the opposite of an event. If we know the probability of an event happening, the complement rule can tell us the probability of it not happening. Here's how it works:

If the probability of an event A happening is denoted by \( P(A) \), then the probability of it not happening \( P(A') \) is given by:
  • \( P(A') = 1 - P(A) \)
In our scenario, we use the complement rule to find the probability that a customer does not buy any extended warranties for both washers and dryers. This probability helps us understand consumer behavior patterns when it comes to avoiding such additional costs.
Union of Events
The union of events in probability relates to the likelihood of either or both of two events happening. When we talk about two events, say A and B, the union is expressed as \( A \cup B \). This includes the following possibilities:
  • Event A happens.
  • Event B happens.
  • Both events A and B happen.
To find the probability of the union of two events \( P(A \cup B) \), we use the formula:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

In our exercise, this calculation helps determine the fraction of customers who purchase at least one extended warranty, either for the washer or the dryer, or both.
Appliance Purchase Probabilities
Understanding appliance purchase probabilities involves analyzing the likelihood of certain buying behaviors concerning washers and dryers. These probabilities not only help in consumer behavior research but also aid manufacturers in designing better service packages.

In this exercise, we look at several probabilities:
  • The probability a customer buys an extended warranty for a washer.
  • The probability a customer buys one for a dryer.
  • And interestingly, the combined probability for both or neither warranties.
These insights reveal consumer preferences and can lead to more tailored warranty offers that fit customer needs. It also assists manufacturers in adjusting their strategies to maximize sales and customer satisfaction.

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Most popular questions from this chapter

A study of the impact of seeking a second opinion about a medical condition is described in the paper "Evaluation of Outcomes from a National Patient- Initiated Second-Opinion Program". Based on a review of 6791 patient-initiated second opinions, the paper states the following: "Second opinions often resulted in changes in diagnosis (14.8\%), treatment \((37.4 \%),\) or changes in both \((10.6 \%)\)." Consider the following two events: \(D=\) event that second opinion results in a change in diagnosis \(T=\) event that second opinion results in a change in treatment a. What are the values of \(P(D), P(T),\) and \(P(D \cap T) ?\) b. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(D\) and \(D^{C}\) and rows corresponding to \(T\) and \(T^{C}\). c. What is the probability that a second opinion results in neither a change in diagnosis nor a change in treatment? d. What is the probability that a second opinion results is a change in diagnosis or a change in treatment?

A rental car company offers two options when a car is rented. A renter can choose to pre-purchase gas or not and can also choose to rent a GPS device or not. Suppose that the events \(A=\) event that gas is pre-purchased \(B=\) event that a GPS is rented are independent with \(P(A)=0.20\) and \(P(B)=0.15\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not gas is pre-purchased and rows corresponding to whether or not a GPS is rented. b. Use the table to find \(P(A \cup B)\). Give a long-run relative frequency interpretation of this probability.

Are people more confident in their answers when the answer is actually correct than when it is not? The article "Female Students Less Confident, More Accurate Than Male Counterparts" described a study that measured medical students" confidence and the accuracy of their responses. Participants categorized their confidence levels using either "sure," "feeling lucky," or "no clue" Define the following events: \(C=\) event that a response is conect \(S=\) event that confidence level is "sure" \(L=\) event that confidence level is "feeling lucky" \(N=\) event that confidence level is "no clue" a. Data from the article were used to estimate the following probabilities for males: $$ \begin{array}{rrr} P(S)=0.442 & P(L)=0.422 & P(N)=0.136 \\ P(C \mid S)=0.783 & P(C \mid L)=0.498 & P(C \mid N)=0.320 \end{array} $$ Use the given probabilities to construct a hypothetical 1000 table with rows corresponding to confidence level and columns comesponding to whether the response was correct or not. b. Calculate the probability that a male student's confidence level is "sure" given that the response is correct. c. Calculate the probability that a male student's confidence level is "no clue" given that the response is incorrect. d. Calculate the probability that a male student's response is correct. e. Data from the article were also used to estimate the following probabilities for females: $$ \begin{array}{rrr} P(S)=0.395 & P(L)=0.444 & P(N)=0.161 \\ P(C 1 S)=0.805 & P(C \mid L)=0.535 & P(C \mid N)=0.320 \end{array} $$ Use the given probabilities to construct a hypothetical 1000 table with rows corresponding to confidence level and columns corresponding to whether the response was comect or not. f. Calculate the probability that a female student's confidence level is "sure" given that the response is correct. g. Calculate the probability that a female student's confidence level is "no clue" given that the response is incorrect. h. Calculate the probability that a female student's response is correct. i. Do the given probabilities and the probabilities that you calculated support the statement in the title of the article? Explain.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

In a particular state, automobiles that are more than 10 years old must pass a vehicle inspection in order to be registered. This state reports the probability that a car more than 10 years old will fail the vehicle inspection is 0.09 . Give a relative frequency interpretation of this probability.
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