/*! 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 51 Consider a system consisting of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=.9\), \(P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The \(1-2\) subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

Short Answer

Expert verified
The probability of the 1-2 and 3-4 subsystem working is 0.81. The system doesn't work with probability 0.0361 and works with probability 0.9639. If a 5-6 subsystem is added, it works with probability 0.999. If there were three components in series in each subsystem, the system would work with probability 0.948

Step by step solution

01

Calculate the probability of subsystem 1-2 and 3-4 working

Components 1 and 2 form a series system. So, both have to work for the subsystem to work. The probability of the subsystem working, denoted as \(P_{1-2}\) is simply the product of the individual probabilities of the component working: \(P_{1-2} = P(1) \cdot P(2) = 0.9 \cdot 0.9 = 0.81\). Similarly for subsystem 3-4, \(P_{3-4} = P(3) \cdot P(4) = 0.9 \cdot 0.9 = 0.81\)
02

Calculate the probability of both subsystems not working

If a subsystem works with probability 0.81, then it does not work with probability \(1 - 0.81 = 0.19\). The system doesn't work when both subsystems don't work. Since these events are independent, their joint probability is the product of the individual probabilities, i.e., \(0.19 \cdot 0.19 = 0.0361\)
03

Calculate the probability of the system working

If the system doesn't work with probability 0.0361, then it works with probability \(1 - 0.0361 = 0.9639\)
04

Calculate probability if a 5-6 subsystem were added

If a subsystem 5-6 were added in parallel with the other two, it would also be a series system, so it would work with the same probability, 0.81. The system as a whole would work if at least one of the subsystems works. Using the principles of probability, the new system would work with probability \(1 - (0.19)^3 = 0.999\)
05

Calculate probability if there were three components in series in each of the two subsystems

If there were three components in series in each subsystem, the probability that one subsystem works would be \(0.9 \cdot 0.9 \cdot 0.9 = 0.729\). The system would then work with probability \(1 - (1-0.729)^2 = 0.948\)

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 91Ó°ÊÓ!

Key Concepts

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

Series and Parallel Subsystems
Understanding the behavior of systems composed of both series and parallel subsystems is critical in fields such as engineering and reliability studies. In a series system, components are arranged in a sequence and the system will work only if all components work. On the other hand, parallel systems are designed with redundancy in mind—where the system will work if at least one component functions.

Let's consider a practical example. Perhaps you have a string of holiday lights (a series system), where if one light bulb fails, the entire string goes dark. Now, imagine having multiple strings connected to each other (a parallel system). In this case, even if one string of lights goes out, the others could still illuminate.

This concept applies to our exercise where Components 1 and 2, and 3 and 4, each form series subsystems. They are, in turn, connected in a parallel arrangement. The system will work as long as at least one of the series subsystems is functional. Such setups are common in safety-critical systems to enhance reliability.
Independent Events in Probability
In probability theory, independence is a fundamental concept. Two events are independent if the occurrence of one event does not influence the probability of the other. For example, when flipping a coin, the result of one flip does not affect the outcome of the next flip—each flip is an independent event.

In the context of our system, Components 1, 2, 3, and 4 are said to work independently. This implies that the functioning of any one component has no bearing on the operation of the others. This characteristic of independence is critical when determining the joint probabilities of the subsystems functioning, as seen in our exercise. Independent events simplify calculations because the joint probability of independent events occurring together is the product of their individual probabilities.
Joint Probability Calculation
Joint probability refers to the likelihood of two or more events occurring simultaneously. To compute the joint probability of independent events, we multiply the probabilities of the individual events. This rule applies when the events do not affect each other—a perfect scenario for our example with subsystems formed by Components 1, 2, 3, and 4.

In the solution to our exercise, we calculated the probability that both subsystems 1-2 and 3-4 would not work by multiplying their respective probabilities of failure. The beauty of independent components within these subsystems is seen here; their joint failure (or success) rates can be easily computed without complex dependencies complicating things. This helps in designing systems with predictable behavior.
Complementary Probability
Complementary probability deals with the concept of 'the other side of the coin' in the probability space. It is used to find the probability of the opposite of a given event. The complementary rule states that the probability of an event not occurring is 1 minus the probability of the event occurring.

Applied to our system's components, if the probability of a component working is 0.9, then the complementary probability—that it does not work—is 1 - 0.9, or 0.1. We see this calculation in the second step of finding the probability that both subsystems 1-2 and 3-4 do not work. The ease of using complementary probability lies in its simplification of problems, especially when it is more straightforward to calculate the chance of an event not happening rather than happening.

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 article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for \(\mathrm{A}\), a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3, P(\) B wins game \()=.2\), and \(P(\) draw \()=.5 .\) Each time a player wins a game, he earns 1 point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that A wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation to estimate \(P(\) A wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than three licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, 9 are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as many of the requests as possible. For example, the city might fill requests for \(2,3,1\), and 3 licenses and then select a request for \(3 .\) Because there is only one license left, the last request selected would receive a license, but only one. a. An individual has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

A single-elimination tournament with four players is to be held. In Game 1 , the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: \(P(\) seed 1 defeats seed 4\()=.8\) \(P(\) seed 1 defeats seed 2\()=.6\) \(P(\) seed 1 defeats seed 3\()=.7\) \(P(\) seed 2 defeats seed 3\()=.6\) \(P(\) seed 2 defeats seed 4\()=.7\) \(P(\) seed 3 defeats seed 4\()=.6\) a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate Game 3 in the tournament? (This will depend on the outcomes of Games 1 and 2.) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

A theater complex is currently showing four R-rated movies, three \(\mathrm{PG}-13\) movies, two \(\mathrm{PG}\) movies, and one \(\mathrm{G}\) movie. The following table gives the number of people at the first showing of each movie on a certain Saturday: $$ \begin{array}{rlc} \text { Theater } & \text { Rating } & \begin{array}{l} \text { Number of } \\ \text { Viewers } \end{array} \\ \hline 1 & \mathrm{R} & 600 \\ 2 & \mathrm{PG}-13 & 420 \\ 3 & \mathrm{PG}-13 & 323 \\ 4 & \mathrm{R} & 196 \\ 5 & \mathrm{G} & 254 \\ 6 & \mathrm{PG} & 179 \\ 7 & \mathrm{PG}-13 & 114 \\ 8 & \mathrm{R} & 205 \\ 9 & \mathrm{R} & 139 \\ 10 & \mathrm{PG} & 87 \\ \hline \end{array} $$Suppose that a single one of these viewers is randomly selected. a. What is the probability that the selected individual saw a PG movie? b. What is the probability that the selected individual saw a PG or a PG-13 movie? c. What is the probability that the selected individual did not see an R movie?

See all solutions

Recommended explanations on Math Textbooks

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.