/*! 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} Q80E Consider the system of component... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q80E (page 90)

Consider the system of components connected as in the accompanying picture. Components \({\rm{1}}\) and \({\rm{2}}\) are connected in parallel, so that subsystem works iff either 1 or 2 works; since \({\rm{3}}\)and\({\rm{4}}\) are connected in series, that subsystem works iff both\({\rm{3}}\)and\({\rm{4}}\)work. If components work independently of one another and P(component i works) \({\rm{ = }}{\rm{.9}}\)for \({\rm{i = }}{\rm{.1,2}}\)and \({\rm{ = }}{\rm{.8}}\)for \({\rm{i = 3,4}}\),calculate P(system works).

Short Answer

Expert verified

The system works is \(\begin{array}{c}{\rm{P( system works ) = 0}}{\rm{.9964}}\\{\rm{ = 99}}{\rm{.64\% }}\end{array}\)

Step by step solution

01

Definition of Independence

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

02

Given parameters

Given: Five valves

\(\begin{array}{l}{\rm{P( component 1 works ) = 0}}{\rm{.9}}\\{\rm{P( component 2 works ) = 0}}{\rm{.9}}\\{\rm{P( component 3 works ) = 0}}{\rm{.8}}\\{\rm{P( component 4 works ) = 0}}{\rm{.8}}\end{array}\)

The components are self-contained.

03

Finding the system works

For independent events, use the following multiplication rule:

\({\rm{P(A and B) = P(A)*P(B)}}\)

For any two events, the following is the general addition rule:

\({\rm{P(A or B) = P(A) + P(B) - P(A and B)}}\)

Use the multiplication rule for independent events:

\({\rm{P( component 1 and 2 work ) = P( component 1 works )*P( component 2 works )}}\)\({\rm{ = 0}}{\rm{.9*0}}{\rm{.9 = 0}}{\rm{.81}}\)

If component \({\rm{1}}\)or component \({\rm{2}}\)work, the top subsystem will also work. For any two occurrences, apply the general addition rule:

\({\rm{P}}\left( {{\rm{top sub system works }}} \right){\rm{ = P( component\;works ) + P( component\;works ) - P(component\;andwork )}}\)\({\rm{ = 0}}{\rm{.9 + 0}}{\rm{.9 - 0}}{\rm{.81 = 0}}{\rm{.99}}\)

If both components \({\rm{3}}\) and \({\rm{4}}\)are operational, the bottom subsystem will function. For independent events, use the multiplication rule:

04

Explanation of the solution

\({\rm{P}}\left( {{\rm{bottom subsystem works }}} \right){\rm{ = P( component\;works )*P( component\;works)}}\)\({\rm{ = 0}}{\rm{.8*0}}{\rm{.8 = 0}}{\rm{.64}}\)

The two subsystems are independent, because each component is independent. Use the multiplication rule for independent events:

\({\rm{P}}\left( {{\rm{both subsystems work}}} \right){\rm{ = P}}\left( {{\rm{top subsystem works}}} \right){\rm{* P}}\left( {{\rm{bottom subsystem works}}} \right)\)\({\rm{ = 0}}{\rm{.99*0}}{\rm{.64 = 0}}{\rm{.6336}}\)

The system will then work if either the top subsystem works or the bottom subsystem works. Use the general addition rule for any two events:

\({\rm{P}}\left( {{\rm{system works}}} \right){\rm{ = P}}\left( {{\rm{top subsystem works}}} \right){\rm{ + P}}\left( {{\rm{bottom subsystem works}}} \right)\)

\(\begin{array}{c}{\rm{ = 0}}{\rm{.99 + 0}}{\rm{.64 - 0}}{\rm{.6336}}\\{\rm{ = 0}}{\rm{.9964 = 99}}{\rm{.64\% }}\end{array}\)

Therefore,The system works is \(\begin{array}{c}{\rm{P( system works ) = 0}}{\rm{.9964}}\\{\rm{ = 99}}{\rm{.64\% }}\end{array}\)

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!

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

An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.

a. If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible?

b. How many experimental runs are there that involve use of the lowest temperature and two lowest pressures?

c. Suppose that five different experimental runs are to be made on the first day of experimentation. If the five are randomly selected from among all the possibilities, so that any group of five has the same probability of selection, what is the probability that a different catalyst is used on each run?

Each of a sample of four home mortgages is classified as fixed rate (F) or variable rate (V).

a. What are the 16 outcomes in S?

b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate?

c. Which outcomes are in the event that all four mortgages are of the same type?

d. Which outcomes are in the event that at most one of the four is a variable-rate mortgage?

e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events?

f. What are the union and intersection of the two events in parts (b) and (c)?

a. A lumber company has just taken delivery on a shipment of \({\rm{10,000 2 \times 4}}\)boards. Suppose that 20% of these boards (\({\rm{2000}}\)) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A 5 {the first board is green} and B 5 {the second board is green}. Compute \({\rm{P(A),P(B)}}\), and \({\rm{P(A}} \cap {\rm{B)}}\) (a tree diagram might help). Are \({\rm{A}}\) and \({\rm{B}}\) independent?

b. With \({\rm{A}}\) and \({\rm{B}}\) independent and \({\rm{P(A) = P(B) = }}{\rm{.2,}}\) what is \({\rm{P(A}} \cap {\rm{B)}}\)? How much difference is there between this answer and \({\rm{P(A}} \cap {\rm{B)}}\)part (a)? For purposes of calculating \({\rm{P(A}} \cap {\rm{B)}}\), can we assume that \({\rm{A}}\)and \({\rm{B}}\) of part (a) are independent to obtain essentially the correct probability?

c. Suppose the shipment consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \({\rm{P(A}} \cap {\rm{B)}}\)? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \({\rm{P(A}} \cap {\rm{B)}}\)?

Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects \({\rm{90\% }}\)of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on \({\rm{20\% }}\)of all defective components. What is the probability that the following occur?

a. A defective component will be detected only by the first inspector? By exactly one of the two inspectors?

b. All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?

Reconsider the system defect situation described in Exercise.

a. Given that the system has a type \(1\) defect, what is the probability that it has a type \({\bf{2}}\) defect?

b. Given that the system has a type \(1\) defect, what is the probability that it has all three types of defects?

c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect?

d. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

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.