Question

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

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

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, \(P_{3-4} = P(3) \cdot P(4) = 0.9 \cdot 0.9 = 0.81\)

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., \(0.19 \cdot 0.19 = 0.0361\)

Calculate the probability of the system working

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

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 \(1 - (0.19)^3 = 0.999\)

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

Key concepts

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.