/*! 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 25 The U.S. Census Bureau provides ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 25

The U.S. Census Bureau provides data on the number of young adults, ages \(18-24,\) who are living in their parents' home. \(^{1}\) Let \(M=\) the event a male young adult is living in his parents' home \(F=\) the event a female young adult is living in her parents' home If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude \(P(M)=.56\) and \(P(F)=.42\) (The World Almanac, 2006 ). The probability that both are living in their parents' home is .24 a. What is the probability at least one of the two young adults selected is living in his or her parents' home? b. What is the probability both young adults selected are living on their own (neither is living in their parents' home)?

Short Answer

Expert verified
a. 0.74 b. 0.26

Step by step solution

01

Identify Given Information

We know that the probability a male young adult lives at home is \(P(M) = 0.56\) and for a female young adult, \(P(F) = 0.42\). Additionally, the probability that both are living at home is \(P(M \cap F) = 0.24\).
02

Calculate Probability of At Least One Living at Home

To find the probability that at least one of the young adults is living at home, we use the formula for the probability of the union of two events: \(P(M \cup F) = P(M) + P(F) - P(M \cap F)\). Substitute the given values: \(P(M \cup F) = 0.56 + 0.42 - 0.24 = 0.74\).
03

Calculate Probability Both Live on Their Own

To find the probability both live on their own (neither is in parents' home), we first find the probability of the complement of \(P(M \cup F)\): \(P((M \cup F)^c) = 1 - P(M \cup F) = 1 - 0.74 = 0.26\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Complement Rule
The complement rule is a fundamental concept in probability that helps us determine the likelihood of an event not happening. Let's break it down to understand it better. In probability, when we have an event, its complement is everything that is not part of that event. For example, if we have an event A with a certain probability, the probability of the complement of A (denoted as \( A^c \)) is the probability that A does not happen. Mathematically, this is expressed as:\[P(A^c) = 1 - P(A)\]In simple terms, if the probability of an event happening is known, say 0.74, then the probability of it not happening, or its complement, would be 1 minus 0.74, which is 0.26. This rule is useful when calculating how likely it is for an event to not occur or determining probabilities of events based on other known information.
  • Use this rule to find the probability the event does not occur.
  • It simplifies calculations when the event probability is known.
Union of Events
The union of events in probability refers to a situation where either event A or event B occurs, or possibly both. The union is denoted by \( A \cup B \). Calculating the probability for the union of two events is straightforward, thanks to the formula:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This formula considers the probability of each event happening and subtracts the probability of both events occurring together (which is counted twice if just added). For the young adult example, we calculated the probability of at least one adult living at home by adding the probabilities of each independently (0.56 and 0.42) and subtracting the probability of both events happening together (0.24), resulting in 0.74.
Calculating the union helps answer questions like "what is the probability of either or both events occurring?"
  • Gives insight into the likeliness of multiple events happening.
  • Useful for "at least one" type probability queries.
Probability of Independent Events
Independent events are two events where the occurrence of one does not affect the occurrence of the other. In our exercise, determining whether events like these, such as whether a male and a female are living at home, can be crucial. If they are independent, the probability of both occurring simultaneously would simply be the product of their individual probabilities:\[P(A \cap B) = P(A) \times P(B)\]However, in this exercise, the probability we calculated for both living at home (0.24) is given directly. This does not follow from multiplying the independent probabilities 0.56 and 0.42 (which would be 0.2352), suggesting there might be some dependence. It's essential to identify whether events are independent to decide how to calculate such combined probabilities correctly.
  • Use the independence rule when events do not influence each other.
  • If direct probability is given, it can indicate dependence or specific conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Visa Card USA studied how frequently young consumers, ages 18 to \(24,\) use plastic (debit and credit) cards in making purchases (Associated Press, January 16,2006 ). The results of the study provided the following probabilities. \(\bullet\)The probability that a consumer uses a plastic card when making a purchase is .37. \(\bullet\)Given that the consumer uses a plastic card, there is a .19 probability that the consumer is 18 to 24 years old. \(\bullet\)Given that the consumer uses a plastic card, there is a 81 probability that the consumer is more than 24 years old. U.S. Census Bureau data show that \(14 \%\) of the consumer population is 18 to 24 years old. a. Given the consumer is 18 to 24 years old, what is the probability that the consumer uses a plastic card? b. Given the consumer is over 24 years old, what is the probability that the consumer uses a plastic card? c. What is the interpretation of the probabilities shown in parts (a) and (b)? d. Should companies such as Visa, MasterCard, and Discover make plastic cards available to the 18 to 24 year old age group before these consumers have had time to establish a credit history? If no, why? If yes, what restrictions might the companies place on this age group?

A survey of magazine subscribers showed that \(45.8 \%\) rented a car during the past 12 months for business reasons, \(54 \%\) rented a car during the past 12 months for personal reasons, and \(30 \%\) rented a car during the past 12 months for both business and personal reasons. a. What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? b. What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?

An experiment has three steps with three outcomes possible for the first step, two outcomes possible for the second step, and four outcomes possible for the third step. How many experimental outcomes exist for the entire experiment?

Students in grades 3 through 8 in New York State are required to take a state mathematics exam. To meet the state's proficiency standards, a student must demonstrate an understanding of the mathematics expected at his or her grade level. The following data show the number of students tested in the New York City school system for grades 3 through 8 and the number who met and did not meet the proficiency standards on the exam (New York City Department of Education website, January 16,2010 ). a. Develop a joint probability table for these data. b. What are the marginal probabilities? What do they tell about the probabilities of meeting or not meeting the proficiency standards on the exam? c. If a randomly selected student is a third grader, what is the probability that the student met the proficiency standards? If the student is a fourth grader, what is the probability that the student met the proficiency standards? d. If a randomly selected student is known to have met the proficiency standards on the exam, what it the probability that the student is a third grader? What is the probability if the student is a fourth grader?

Assume that we have two events, \(A\) and \(B\), that are mutually exclusive. Assume further that we know \(P(A)=.30\) and \(P(B)=.40\). a. What is \(P(A \cap B) ?\) b. What is \(P(A | B) ?\) c. \(\quad\) A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Do you agree with this statement? Use the probability information in this problem to justify your answer. d. What general conclusion would you make about mutually exclusive and independent events given the results of this problem?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.