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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 \(1 - (1-0.729)^2 = 0.948\)

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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.

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Most popular questions from this chapter

A transmitter is sending a message using a binary code, namely, a sequence of 0's and 1's. Each transmitted bit \((0\) or 1\()\) must pass through three relays to reach the receiver. At each relay, the probability is \(.20\) that the bit sent on is different from the bit received (a reversal). Assume that the relays operate independently of one another: transmitter \(\rightarrow\) relay \(1 \rightarrow\) relay \(2 \rightarrow\) relay \(3 \rightarrow\) receiver a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent on by all three relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? (Hint: The eight experimental outcomes can be displayed on a tree diagram with three generations of branches, one generation for each relay.)

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\) and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F ?\) b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F) ?\) A Venn diagram might help.)

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lr}\text { Money market } & 20 \% \\ \text { Short-term bond } & 15 \% \\ \text { Intermediate-term bond } & 10 \% \\ \text { Long-term bond } & 5 \% \\ \text { High-risk stock } & 18 \% \\ \text { Moderate-risk stock } & 25 \% \\ \text { Balanced fund } & 7 \%\end{array}\) A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

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