/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Consider the type of clothes dry... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the type of clothes dryer (gas or electric) purchased by each of five different customers at a certain store. a. If the probability that at most one of these purchases an electric dryer is \(.428\), what is the probability that at least two purchase an electric dryer? b. If \(P\) (all five purchase gas) \(=.116\) and \(P\) (all five purchase electric) \(=.005\), what is the probability that at least one of each type is purchased?

Short Answer

Expert verified
a) 0.572 b) 0.879

Step by step solution

01

Understand the Complement Rule

The probability that at least two customers purchase an electric dryer is the complement of the probability that at most one customer purchases an electric dryer. Mathematically, if \( A \) is the event that at least two purchase, then \( P(A) = 1 - P(\text{at most one} \)].
02

Calculate the Complementary Probability

Given that \( P(\text{at most one electric}) = 0.428 \), the probability that at least two purchase electric dryers is \( P(A) = 1 - 0.428 \).
03

Compute Probability for At Least One of Each Type

To solve part b, recognize that the probability of at least one of each type is the complement of the probability that all purchases are either gas or electric. Since \( P(\text{all gas}) = 0.116 \) and \( P(\text{all electric}) = 0.005 \), use \( P(\text{at least one of each type}) = 1 - [P(\text{all gas}) + P(\text{all electric})] \).
04

Apply Complementary Calculations

Calculate \( P(\text{at least two purchase electric}) = 1 - 0.428 = 0.572 \). For part b, calculate \( P(\text{at least one of each type}) = 1 - (0.116 + 0.005) = 0.879 \).
05

Interpret the Results

The results tell us the likelihood that at least two customers choose electric dryers, and that at least one of each type of dryer is purchased. Specifically, the probability that at least two buy electric is \( 0.572 \), and the probability that at least one customer purchases each type of dryer is \( 0.879 \).

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Key Concepts

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

Complement Rule
The concept of the Complement Rule is a fundamental part of understanding probability. It helps simplify what may seem complex at first. When calculating probabilities, instead of figuring out the chance of something happening directly, we can determine its complement, which is often simpler. For this rule, the probability of an event not happening can be subtracted from 1 to get the probability of the event occurring.

In the context of this exercise, figuring out the probability that at least two customers purchased an electric dryer becomes more manageable by looking at its complement. Instead of calculating directly, we first determine the probability that at most one customer buys an electric dryer, which is already given as 0.428. Then, using the Complement Rule, we find the probability of at least two buying an electric dryer as 1 minus 0.428, equaling 0.572.
Electric Dryer
Electric dryers are a popular household appliance choice for many, known for their convenience and the fact that they usually just require an electrical outlet without the need for a gas connection. They operate by drawing air in and then passing it over a heating element, which is powered by electricity.

In probability exercises like this, electric dryers serve as an example of variety in choice. The focus is on determining how often households choose one type of dryer over another. By understanding the purchasing habits—like whether at least two customers choose electric dryers—we apply probability calculations to real-life scenarios. This gives insights that go beyond just theory and predict consumer behavior.
Gas Dryer
Gas dryers function differently from electric dryers, as they use natural gas or propane to create heat. Many users prefer them for their cost efficiency and faster drying times. Unlike electric dryers, they require a dedicated gas line, which may influence purchasing decisions.

In analyzing customer purchases, understanding the likelihood of choosing a gas dryer complements the examination of probability within the exercise. Calculating probabilities around gas dryer purchases helps in hypotheses such as the chance of all customers choosing gas or observing a mix of gas and electric dryers among purchases. This understanding can guide retailers in planning inventory and marketing strategies.
Probability Calculations
Probability calculations are essential in predicting how likely specific outcomes are, especially when various options exist. These calculations are based on the understanding of all possible outcomes and how they contribute to a desired event.

In the given problem, probability calculations involve not just one straightforward estimation, but a series of complementary assessments. First, by understanding the probability that all five customers choose electric or all choose gas, and using their complements, we deduce the scenario where there's at least one of each type: gas and electric. Using these calculations, we arrive at the solution: there's an 87.9% probability of having a mix of purchases.

Engaging with probability calculations in exercises gives insight into decision making under uncertainty, a skill that's valuable in numerous real-world applications.

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Most popular questions from this chapter

A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let \(A\) denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5, P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

A certain company sends \(40 \%\) of its overnight mail parcels via express mail service \(E_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (denote the event "late delivery" by \(L\) ). If a record of an overnight mailing is randomly selected from the company's file, what is the probability that the parcel went via \(E_{1}\) and was late?

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is \(.05\). What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly? Answer this question for 25 successive points.

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+P\left(A^{\prime} \mid B\right)=1\).

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

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