/*! 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 140 A trimotor plane has three engin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 140

A trimotor plane has three engines-a central engine and an engine on each wing. The plane will crash only if the central engine fails and at least one of the two wing engines fails. The probability of failure during any given flight is \(.005\) for the central engine and \(.008\) for each of the wing engines. Assuming that the three engines operate independently, what is the probability that the plane will crash during a flight?

Short Answer

Expert verified
The probability that the plane will crash during a flight is 0.00008.

Step by step solution

01

Identify the probability of each event

We are given the probabilities of failure for each engine: \(P(Central) = 0.005\) for the central engine failure and \(P(Wing) = 0.008\) for each wing engine failure. Because both wing engines have the same failure rate, we also know that \(P(Wing1) = P(Wing2) = 0.008\).
02

Calculate the probability of independent events

We know that the plane will crash if both the central engine fails and at least one of the wing engines fail. These are independent events, so we calculate the joint probability by multiplying the probabilities of each event. Our scenarios are either the central and the first wing engine fail, or the central and the second wing engine fail. Therefore, the total probability of crashing is given by: \(P(Crash) = P(Central \cap Wing1) + P(Central \cap Wing2) = P(Central) \cdot P(Wing1) + P(Central) \cdot P(Wing2) = 0.005 \cdot 0.008 + 0.005 \cdot 0.008 = 0.00004 + 0.00004 \).
03

Evaluate the final probability

Adding the probabilities calculated in Step 2 we get the final probability of the plane crashing. Thus, \(P(Crash) = 0.00004 + 0.00004 = 0.00008\).

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.

Independent Events
Understanding independent events is quite important when dealing with probabilities. Independent events are those in which the result or occurrence of one event does not affect the occurrence of another. In other words, knowing the outcome of one event gives no information about another.
In our exercise, each of the three engines operates independently. This implies that whether the central engine fails has no effect on whether the wing engines fail, and vice versa. This independence is crucial because it allows us to calculate joint probabilities by simply multiplying the probability of individual events. This concept is foundational in probability theory and aids in simplifying complex probability calculations.
Joint Probability
Calculating joint probabilities is essential when we are interested in determining the likelihood of multiple events occurring together. For independent events, like the engines in our problem, the joint probability of two events occurring simultaneously is simply the product of their individual probabilities.
To break it down: the plane will crash if the central engine fails and at least one of the wing engines fails. The joint probability approach allows us to find this by multiplying:
  • The probability of the central engine failing,
  • and the probability of one wing engine failing.
For both the first and second wing engines, this results in the formula: \[ P(Crash) = P(Central \cap Wing1) + P(Central \cap Wing2) = 0.005 \times 0.008 + 0.005 \times 0.008 \].
Utilizing joint probability gives a clear picture of compound events' likelihood.
Engine Failure
Engine failure in aviation is a critical topic lurking in the backdrop of flight safety. In this exercise, we are given specific failure probabilities for each engine. The central engine has a failure probability of 0.005, while each wing engine has a probability of 0.008. These small probabilities reflect the reliability and safety standards in aviation.
Engine failure considerations are central to designing robust systems in aircraft. The exercise also reminds us about the importance of independent systems in reducing the likelihood of a catastrophic event like a plane crash. By keeping engines independent, the aircraft can keep functioning even if one engine has issues. In practical terms, this redundancy protects against total system failure, aiming to make flights as safe as possible.

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 consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$\begin{array}{cccc} \hline & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} \mathbf{3 0} \text { Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \\ \hline \end{array}$$ a. Suppose one flight is selected at random from these 1700 flights. Find the following probabilities. i. \(P(\) more than 1 hour late and airline \(\mathrm{A}\) ) ii. \(P(\) airline \(\mathrm{B}\) and less than 30 minutes late) b. Find the joint probability of events " 30 minutes to 1 hour late" and "more than 1 hour late." Is this probability zero? Explain why or why not.

Five hundred employees were selected from a city's large private companies, and they were asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way classification table was prepared. $$\begin{array}{lcc} \hline & \text { Yes } & \text { No } \\ \hline \text { Men } & 225 & 75 \\ \text { Women } & 150 & 50 \end{array}$$ a. If one employee is selected at random from these 500 employees, find the probability that this employee i. is a woman ii. has retirement benefits iii. has retirement benefits given the employee is a man iv. is a woman given that she does not have retirement benefits b. Are the events "man" and "yes" mutually exclusive? What about the events "yes" and "no?" Why or why not? c. Are the events "woman" and "yes" independent? Why or why not?

A consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$\begin{array}{lccc} & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} 30 \text { Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \end{array}$$ If one flight is selected at random from these 1700 flights, find the following probabilities. a. \(P(\) more than 1 hour late or airline \(A\) ) b. \(P(\) airline \(\mathrm{B}\) or less than 30 minutes late) c. \(P(\) airline A or airline \(\mathrm{B}\) )

The probability that a randomly selected college student attended at least one major league baseball game last year is .12. What is the complementary event? What is the probability of this complementary event?

An appliance repair company that makes service calls to customers' homes has found that \(5 \%\) of the time there is nothing wrong with the appliance and the problem is due to customer error (appliance unplugged, controls improperly set, etc.). Two service calls are selected at random, and it is observed whether or not the problem is due to customer error. Draw a tree diagram. Find the probability that in this sample of two service calls a. both problems are due to customer error b. at least one problem is not due to customer error
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.