/*! 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 19 A consumer organization estimate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 19

A consumer organization estimates that over a 1-year period \(17 \%\) of cars will need to be repaired once, \(7 \%\) will need repairs twice, and \(4 \%\) will require three or more repairs. What is the probability that a car chosen at random will need a) no repairs? b) no more than one repair? c) some repairs?

Short Answer

Expert verified
a) 72%; b) 89%; c) 28%

Step by step solution

01

Understand Given Information

We are given the probabilities for the number of times a car needs repairs over a year. Specifically, the probabilities provided are: 17% of cars need one repair, 7% need two repairs, and 4% need three or more repairs. These probabilities add up to the cars needing repairs.
02

Calculate Probability of No Repairs (Part a)

To find the probability that a car will not need any repairs, we must consider that either it has 0 repairs or any of the other mentioned options. The total probability must be 100%. Hence, if 28% of cars need some form of repair, then the probability of no repairs is the complement of this. Calculate it as follows: \[ P(0 \, \text{repairs}) = 1 - (0.17 + 0.07 + 0.04) = 1 - 0.28 = 0.72 \] So, there is a 72% chance that a car won't need any repairs.
03

Probability of No More Than One Repair (Part b)

For this part, we add the probabilities of cars needing either zero or one repair. We already found in the previous step that the probability of zero repairs is 72%. The probability of exactly one repair is 17%.Calculate it as follows:\[ P(\text{no more than one repair}) = P(0 \, \text{repairs}) + P(1 \, \text{repair}) = 0.72 + 0.17 = 0.89 \]Thus, there is an 89% chance that a car will need no more than one repair.
04

Probability of Some Repairs (Part c)

To find the probability that a car will need some repairs, we consider cars needing one, two, or three or more repairs. Utilizing earlier calculations, we can subtract the probability of zero repairs from 1:\[ P(\text{some repairs}) = 1 - P(0 \, \text{repairs}) = 1 - 0.72 = 0.28 \]There is a 28% chance that a car will need some repairs.

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.

Complementary Probability
Understanding complementary probability can simplify complex probability problems. This concept revolves around two events that are mutually exclusive and exhaustive. That means they do not happen at the same time, and together they cover all possibilities. In probability terms, if you have an event \(A\), its complement \(A'\) represents everything that is not part of \(A\). The probability of event \(A\) and its complement adds up to 1.
Applying this to our original exercise, determining the probability of zero repairs involves finding the complement of cars needing one or more repairs. Since 28% of cars need repairs at least once, the complementary probability that they will need no repairs is the remaining percentage: \(1 - 0.28 = 0.72\). Complementary probability allows you to calculate probabilities by subtraction when direct calculation seems difficult or cumbersome.
Probability Calculations
Probability calculations require adding probabilities for various mutually exclusive events. When breaking down our exercise, the total probability is essentially the sum of all potential scenarios within the experiment.
  • Single Event Probability: Understanding basic scenarios involves finding the specific probability of each event, like the 17% probability of cars needing one repair.
  • Combined Probability: To find scenarios involving multiple possibilities, add up the individual probabilities, such as combining zero and one repair events. In our example, this sums up to 89% for zero or one repair.
Seamlessly blending these probabilities gives insight into the likelihood of complete outcomes, accommodating for overlap effectively. Always ensure the total probability equals 1, just as 0.72 for zero repairs plus 0.28 for some repairs confirms.
Real-World Applications of Probability
Probability isn't just a mathematical concept. It is widely used in everyday life, aiding in informed decision-making. This exercise provides insights often applicable in real-world scenarios, such as anticipating car maintenance needs.
  • Risk Assessment: Understanding the probability of repairs helps manufacturers and consumers anticipate future maintenance, leading to cost savings and informed purchasing choices.
  • Resource Allocation: Companies use probability to allocate resources efficiently, like deploying enough workforce to meet expected repair demands based on probabilities.
By applying probability calculations effectively, businesses and individuals can better manage risks and plan strategically. This predictive power demonstrates probability's significant impact beyond theoretical exercises.

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

You shuffle a deck of cards and then start turning them over one at a time. The first one is red. So is the second. And the third. In fact, you are surprised to get 10 red cards in a row. You start thinking, "The next one is due to be black!" a) Are you correct in thinking that there's a higher probability that the next card will be black than red? Explain. b) Is this an example of the Law of Large Numbers? Explain.

According to Pew Research, the contact rate (probability of contacting a selected household) was \(69 \%\) in 1997 and \(76 \%\) in 2003 . However, the cooperation rate (probability of someone at the contacted household agreeing to be interviewed) was \(58 \%\) in 1997 and dropped to \(38 \%\) in 2003 . a) What is the probability (in 2003) of obtaining an interview with the next household on the sample list? (To obtain an interview, an interviewer must both contact the household and then get agreement for the interview.) b) Was it more likely to obtain an interview from a randomly selected household in 1997 or in \(2003 ?\)

You used the Multiplication Rule to calculate probabilities about the Calculus background of your Statistics groupmates in Exercise \(22 .\) a) What must be true about the groups in order to make that approach valid? b) Do you think this assumption is reasonable? Explain.

Funding for many schools comes from taxes based on assessed values of local properties. People's homes are assessed higher if they have extra features such as garages and swimming pools. Assessment records in a certain school district indicate that \(37 \%\) of the homes have garages and \(3 \%\) have swimming pools. The Addition Rule might suggest, then, that \(40 \%\) of residences have a garage or a pool. What's wrong with that reasoning?

After an unusually dry autumn, a radio announcer is heard to say, "Watch out! We'll pay for these sunny days later on this winter." Explain what he's trying to say, and comment on the validity of his reasoning.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.