/*! 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 22 The route used by a certain moto... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 22

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is \(.4\), the analogous probability for the second signal is \(.5\), and the probability that he must stop at at least one of the two signals is .7. What is the probability that he must stop a. At both signals? b. At the first signal but not at the second one? c. At exactly one signal?

Short Answer

Expert verified
a) 0.2, b) 0.2, c) 0.5

Step by step solution

01

Define Probabilities

Let's define the following probabilities: \( P(A) = 0.4 \) (probability of stopping at the first signal), \( P(B) = 0.5 \) (probability of stopping at the second signal), and \( P(A \cup B) = 0.7 \) (probability of stopping at at least one of the signals). We need to find: (a) \( P(A \cap B) \), (b) \( P(A \cap B') \), and (c) \( P(A \cap B') + P(A' \cap B) \).
02

Use Union Formula for Part (a)

We use the formula for the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substitute the known values to find \( P(A \cap B) \).
03

Calculate Intersection Probability

Substitute the values into the formula: \( 0.7 = 0.4 + 0.5 - P(A \cap B) \). This simplifies to \( P(A \cap B) = 1 - 0.7 = 0.2 \). Therefore, the probability that he must stop at both signals is \( P(A \cap B) = 0.2 \).
04

Calculate Probability for Part (b)

The event "stopping at the first signal but not the second" is represented as \( A \cap B' \). The probability is calculated as \( P(A \cap B') = P(A) - P(A \cap B) \).
05

Solve for (b) using Known Values

Substitute the known values: \( P(A \cap B') = 0.4 - 0.2 = 0.2 \). Thus, the probability that he must stop at the first signal but not the second one is \( 0.2 \).
06

Calculate Probability for Part (c)

The probability of stopping at exactly one signal is \( P(A \cap B') + P(A' \cap B) \). We already know \( P(A \cap B') = 0.2 \). Calculate \( P(A' \cap B) = P(B) - P(A \cap B) \).
07

Solve for Remaining Probability

Substitute the known values: \( P(A' \cap B) = 0.5 - 0.2 = 0.3 \). Therefore, \( P(A \cap B') + P(A' \cap B) = 0.2 + 0.3 = 0.5 \).
08

Summary of Results

The probability of stopping at both signals is 0.2, at the first but not the second is 0.2, and at exactly one signal is 0.5.

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.

Intersection Probability
Intersection probability takes us into the realm of finding the likelihood of two events occurring simultaneously. For instance, if we want the probability that the motorist stops at both traffic signals, we use the notation \( P(A \cap B) \). In the given problem, we leverage the union formula to find this probability. It essentially tells us how to overlap these two individual probabilities, \( P(A) \) and \( P(B) \), minus the overlap or common part, \( P(A \cap B) \). Through calculation, the intersection probability is found to be \( 0.2 \), meaning there is a 20% chance the motorist stops at both signals.

Understanding intersection probability helps in situations where both events could potentially happen together, providing a better grasp on combined event likelihoods.
Event Probability
Event probability refers to the likelihood of a specific event occurring out of all possible outcomes. It is a fundamental concept in probability, denoted as \( P(E) \), where \( E \) is an event. In our scenario, the event probability is expressed as \( P(A) \) for stopping at the first signal and \( P(B) \) for stopping at the second signal. Here, \( P(A) = 0.4 \) and \( P(B) = 0.5 \).

The probability value provides insight into how often we can expect the event to occur when the scenario is repeated multiple times. For instance, a 40% chance translates into the motorist stopping at the first signal, on average, 4 out of every 10 commutes.
Union Formula
The union formula is key when calculating the likelihood of either one or both of the events happening. It's expressed as \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). In this situation, it helps us determine \( P(A \cap B) \), which is the intersection - or both signals in this case - based on the probabilities given and the union probability, \( P(A \cup B) = 0.7 \).

Applying this formula ensures that we don't double-count the scenarios where both events happen simultaneously. It efficiently combines individual event probabilities, adjusting for their overlap.
Conditional Probability
Conditional probability concerns the probability of an event occurring given that another event has already occurred. Although it isn't directly used in the exercise, it's related to our understanding of \( P(A \cap B') \) when the condition is being at the first signal and not the second.

While the exercise focuses more on individual and combined probabilities of independent events, conditional probability can be a natural extension when events influence each other. This concept helps assess probabilities under a given set of conditions, further illustrating how events relate beyond mere occurrence.

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

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that \(16 \%\) of all ticks in a certain location carry Lyme disease, \(10 \%\) carry HGE, and \(10 \%\) of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

Suppose that vehicles taking a particular freeway exit can turn right \((R)\), turn left \((L)\), or go straight \((S)\). Consider observing the direction for each of three successive vehicles. a. List all outcomes in the event \(A\) that all three vehicles go in the same direction. b. List all outcomes in the event \(B\) that all three vehicles take different directions. c. List all outcomes in the event \(C\) that exactly two of the three vehicles turn right. d. List all outcomes in the event \(D\) that exactly two vehicles go in the same direction. e. List outcomes in \(D^{\prime}, C \cup D\), and \(C \cap D\).

A computer consulting firm presently has bids out on three projects. Let \(A_{i}=\\{\) awarded project \(i\\}\), for \(i=1,2,3\), and suppose that \(P\left(A_{1}\right)=.22, P\left(A_{2}\right)=.25, P\left(A_{3}\right)=.28\), \(P\left(A_{1} \cap A_{2}\right)=.11, P\left(A_{1} \cap A_{3}\right)=.05, P\left(A_{2} \cap A_{3}\right)=.07\), \(P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01\). Express in words each of the following events, and compute the probability of each event: a. \(A_{1} \cup A_{2}\) b. \(A_{1}^{\prime} \cap A_{2}^{\prime}\) c. \(A_{1} \cup A_{2} \cup A_{3}\) d. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}\) e. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}\) f. \(\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}\)

According to a July 31,2013 , posting on cnn.com subsequent to the death of a child who bit into a peanut, a 2010 study in the journal Pediatrics found that \(8 \%\) of children younger than 18 in the United States have at least one food allergy. Among those with food allergies, about \(39 \%\) had a history of severe reaction. a. If a child younger than 18 is randomly selected, what is the probability that he or she has at least one food allergy and a history of severe reaction? b. It was also reported that \(30 \%\) of those with an allergy in fact are allergic to multiple foods. If a child younger than 18 is randomly selected, what is the probability that he or she is allergic to multiple foods?

The Reviews editor for a certain scientific journal decides whether the review for any particular book should be short (1-2 pages), medium (3-4 pages), or long (5-6 pages). Data on recent reviews indicates that \(60 \%\) of them are short, \(30 \%\) are medium, and the other \(10 \%\) are long. Reviews are submitted in either Word or LaTeX. For short reviews, \(80 \%\) are in Word, whereas \(50 \%\) of medium reviews are in Word and \(30 \%\) of long reviews are in Word. Suppose a recent review is randomly selected. a. What is the probability that the selected review was submitted in Word format? b. If the selected review was submitted in Word format, what are the posterior probabilities of it being short, medium, or long?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

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.