/*! 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 69 A new sports car model has defec... [FREE SOLUTION] | 91Ó°ÊÓ

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A new sports car model has defective brakes 15 percent of the time and a defective steering mechanism 5 percent of the time. Let's assume (and hope) that these problems occur. independently. If one or the other of these problems is present, the car is called a "lemon:". If both of these problems are present, the car is a "hazard." Your instructor purchased one of these cars yesterday. What is the probability it is: a. A lemon? b. A hazard?

Short Answer

Expert verified
a. 19.25% chance of being a lemon; b. 0.75% chance of being a hazard.

Step by step solution

01

Understand the Given Probabilities

Determine the given probabilities: the probability of having defective brakes is 15% (or 0.15) and the probability of having a defective steering mechanism is 5% (or 0.05). The problems are stated to be independent of each other.
02

Define Events for Lemon and Hazard

Let event A be the car has defective brakes, and event B be the car has a defective steering mechanism. The car is a lemon if at least one of these issues is present, and a hazard if both issues are present.
03

Calculate Probability of a Hazard

Since the events are independent, the probability of both issues (a hazard) occurring is the product of their individual probabilities. \( P(A \cap B) = P(A) \times P(B) = 0.15 \times 0.05 = 0.0075 \).
04

Calculate Probability of a Lemon

A lemon is a car with at least one defect (either brakes or steering or both). Use the formula for the union of two independent events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substitute the known probabilities: \( P(A \cup B) = 0.15 + 0.05 - 0.0075 = 0.1925 \).

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Key Concepts

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

Independent Events
In probability theory, events are considered independent when the occurrence of one event does not influence the outcome of another. This is an essential concept when calculating probabilities for situations where multiple events occur simultaneously.
For instance, in the sports car example, defective brakes and a defective steering mechanism are independent events. This means that the likelihood of one defect happening does not affect the likelihood of the other defect occurring.
Understanding independence is crucial because it allows us to use the product rule for independent events. When events are independent, the probability of both occurring is simply the product of their individual probabilities.
Probability of Union
The probability of a union of two events, often noted as \( P(A \cup B) \), refers to the likelihood of at least one of the events happening. For independent events, this is crucial when we want to know the probability that either one of the events or both occur.
In mathematical terms, the formula is:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
This formula accounts for the fact that the probability of both events occurring (the intersection) is subtracted to avoid double-counting.
Applying this to the sports car scenario, where a lemon is defined as having at least one defect, makes the calculation clear. You add the probabilities of each defect occurring individually and then subtract the probability of both defects happening together.
Probability of Intersection
The probability of intersection, denoted as \( P(A \cap B) \), represents the likelihood of both events occurring simultaneously.
When dealing with independent events, calculating the intersection is straightforward:
  • Multiply the individual probabilities of each event.
For example, in our car problem, the probability that it is a hazard (both brakes and steering are defective) involves multiplying the probability of each defect: 0.15 for brakes and 0.05 for steering. Thus, the intersection probability ensures we see how likely both issues are to occur together. This type of calculation is simple yet a fundamental part of understanding joint probability in scenarios involving multiple risk factors.

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Most popular questions from this chapter

The events \(A\) and \(B\) are mutually exclusive. Suppose \(P(A)=.30\) and \(P(B)=.20 .\) What is the probability of either \(A\) or \(B\) occurring? What is the probability that neither \(A\) nor \(B\) will happen?

The number of times a particular event occurred in the past is divided by the number of occurrences. What is this approach to probability called?

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