/*! 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 45 You bought a new set of four tir... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 45

You bought a new set of four tires from a manufacturer who just announced a recall because \(2 \%\) of those tires are defective. What is the probability that at least one of yours is defective?

Short Answer

Expert verified
The probability that at least one tire is defective is approximately 7.76%.

Step by step solution

01

Understand the Problem

We need to find the probability that at least one of the four tires is defective. This is a classic problem of complementary probability where we first find the probability that none are defective and subtract it from 1.
02

Find the Probability of a Non-Defective Tire

The probability that a single tire is not defective is given by the complement of the probability that it is defective. Since the probability of a tire being defective is \(2\%\) or \(0.02\), the probability that it is not defective is \(1 - 0.02 = 0.98\).
03

Find the Probability of No Defective Tires

Since the probability of one tire being non-defective is \(0.98\), for four independent tires, the probability that all are non-defective is \((0.98)^4\). Calculate this probability: \[(0.98)^4 = 0.98 \times 0.98 \times 0.98 \times 0.98 = 0.92236816.\]
04

Use Complementary Probability

The probability that at least one tire is defective is the complement of the probability that none of the tires are defective. This is calculated as \[1 - (0.98)^4 = 1 - 0.92236816 = 0.07763184.\]
05

Interpret the Result

The probability that at least one of the four tires is defective is approximately \(0.0776\) or \(7.76\%\).

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
Complementary probability is a concept that allows us to find the probability of an event happening by first calculating the probability of the event not happening. This might sound counterintuitive at first, but it is very useful in certain cases.

For example, when you want to find the probability of at least one tire being defective out of four, it's easier to initially calculate the probability that none of them are defective.
  • This is because the calculation for none being defective is straightforward and involves multiplying probabilities for independent events (in this case, each tire).
  • Once you have that, subtracting the result from 1 provides the desired probability of at least one being defective.
Complementary probability is beneficial when facing problems that would otherwise require enumerating complex combinations.
Defective Items
When dealing with defective items, it's important to know the probability of defect. In our exercise, we have tires, and each has a 2% probability of being defective.
  • This is a straightforward percentage that tells us, on average, 2 out of 100 tires are expected to have defects.
  • Understanding this probability is crucial as it directly impacts the calculations you perform to assess the risks involved.
The information about defective items helps you decide whether additional steps or safeguards are required. For example, you might choose to further inspect the tires prior to usage.
Independent Events
Independent events are essential when calculating total probabilities across multiple instances. An event is said to be independent if its outcome doesn't affect the outcome of another.

For example, the defectiveness of each of the four tires is an independent event:
  • The probability of one tire being defective is 2%, regardless of the condition of the others.
  • Because they are independent, you can multiply the probabilities of individual non-defective events to find the total probability of all being non-defective.
In scenarios like this, independence simplifies calculations and reinforces understanding that each event stands alone.
Probability Calculation
Probability calculation can seem daunting, but breaking it down makes it manageable. Here鈥檚 a simple path using our tire example:

First, calculate the probability of a single event happening or not, such as a tire being non-defective:
  • Given a 2% chance of defect, a tire has a 98% chance of being non-defective: \(P( ext{non-defective}) = 1 - 0.02 = 0.98\)
Next, for multiple independent events (4 tires here), use:
  • \(P( ext{all 4 non-defective}) = 0.98 \times 0.98 \times 0.98 \times 0.98 = (0.98)^4 = 0.92236816\)
Finally, apply complementary probability:
  • \(P( ext{at least 1 defective}) = 1 - (0.98)^4 = 1 - 0.92236816 = 0.07763184\)
  • This result tells you there's a 7.76% chance that at least one of the tires is defective.
Understanding these basic calculations enables you to employ probability in real-life decision-making with confidence.

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

A Pew Research poll in 2011 asked 2005 U.S. adults whether being a father today is harder than it was a generation ago. Here's how they responded: $$\begin{array}{lc}\text { Response } & \text { Number } \\ \text { Easier } & 501 \\ \text { Same } & 802 \\ \text { Harder } & 682 \\ \text { No Opinion } & 20 \\ \text { Total } & 2005\end{array}$$ If we select a respondent at random from this sample of 2005 adults, a) what is the probability that the selected person responded "Harder"? b) what is the probability that the person responded the "Same" or "Easier"?

A certain bowler can bowl a strike \(70 \%\) of the time. What's the probability that she a) goes three consecutive frames without a strike? b) makes her first strike in the third frame? c) has at least one strike in the first three frames? d) bowls a perfect game ( 12 consecutive strikes)?

Your list of favorite songs contains 10 rock songs, 7 rap songs, and 3 country songs. a) What is the probability that a randomly played song is a rap song? b) What is the probability that a randomly played song is not country?

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?

Census reports for a city indicate that \(62 \%\) of residents classify themselves as Christian, \(12 \%\) as Jewish, and \(16 \%\) as members of other religions (Muslims, Buddhists, etc.). The remaining residents classify themselves as nonreligious. A polling organization seeking information about public opinions wants to be sure to talk with people holding a variety of religious views, and makes random phone calls. Among the first four people they call, what is the probability they reach a) all Christians? b) no Jews? c) at least one person who is nonreligious?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.