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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 these four outcomes are independent (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 was 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 subsystem working is 0.81, and of it not working is 0.19. The probability of the 3-4 subsystem not working is also 0.19. The probability of the system not working is 0.0361, while it's 0.9639 for it to work. Adding another subsystem in parallel would increase the system's reliability, while adding an extra component in the series to each subsystem would decrease the system's reliability.

Step by step solution

01

Probability of 1-2 subsystem working

The probability that both components of the 1-2 subsystem work is the product of their respective probabilities since they work independently. This is calculated as \(P(1-2) = P(1) \times P(2) = 0.9 \times 0.9 = 0.81) .
02

Probability of 1-2 subsystem not working

The probability that the 1-2 subsystem doesn't work is 1 minus the probability that it works. This can be calculated as \(P'(1-2) = 1 - P(1-2) = 1 - 0.81 = 0.19\) .
03

Probability of 3-4 subsystem not working

Following the same rules as in Step 1 and Step 2, the probability of the 3-4 subsystem not working can be calculated as \(P'(3-4) = 1 - P(3) \times P(4) = 1 - 0.9 \times 0.9 = 0.19\) .
04

Probability of system not working

The system won't work if both the 1-2 and 3-4 subsystems don't work. Since they work independently, this can be calculated as \(P(\text{System Not Working}) = P'(1-2) \times P'(3-4) = 0.19 \times 0.19 = 0.0361\) .
05

Probability of the system working

The probability of the system working is 1 minus the probability that it doesn't work. Hence, \(P(\text{System Working}) = 1 - P(\text{System Not Working}) = 1 - 0.0361 = 0.9639\) .
06

Adding 5-6 subsystem

If a 5-6 subsystem with the same probabilities as other components were added in parallel, it would increase the system's reliability since now the system would work if any of the three subsystems worked. However, exact calculation can't be performed without knowing the values of \(P(5)\) and \(P(6)\).
07

Three components in a subsystem

If there were three components in series in each of the two subsystems, then the subsystems would be less reliable since to work, all three components would need to work. The probability of subsystem working would be now \(P(1) \times P(2) \times P(3)\) and so the system's reliability would decrease.

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Key Concepts

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

Independent Events
In probability, independent events mean that the outcome of one event does not affect the outcome of another. For the problem at hand, the functioning of each component in the system is independent of the others. This independence allows us to calculate the probability of multiple events occurring simultaneously by multiplying their individual probabilities. Imagine flipping a coin twice; each flip doesn't affect the other. Similarly, if Component 1 operates with a probability of 0.9, this probability isn't influenced by whether Component 2 works or not. Hence, their combined functioning is calculated as the product of their probabilities. Thus, independence helps simplify complex systems into manageable parts.
Series and Parallel Systems
Series and parallel systems describe different ways components in a system are connected. In a series system, all components must work for the system to work. It's like old Christmas lights; if one bulb goes out, the whole string may not light. For the given exercise:
  • Components 1 and 2 form a series subsystem. Both must function for the subsystem to work with a combined probability of 0.81.
  • In contrast, parallel systems allow for more reliability since only one of the parallel paths needs to function for the whole system to work. This setup ensures higher reliability, as the given problem combines two series subsystems in parallel: only one series needs to function for the whole system to be operational.
Reliability Engineering
Reliability engineering focuses on ensuring systems continue to perform under expected conditions. In our problem, reliability means estimating how often the system will function correctly. This is crucial for applications like safety-critical systems, where failures might cause severe consequences. Understanding reliability involves:
  • Recognizing the system's configuration (like series or parallel).
  • Calculating the probability of each component's success and failure.
  • Using these calculations to predict overall system performance success or failure rates.
It's all about maximizing efficiency and minimizing downtime, which is why adding more parallel paths (as explored with the hypothetical 5-6 subsystem) enhances reliability.
Probability Calculations
Probability calculations allow us to quantify the likelihood of various outcomes. In reliability engineering, this means determining how likely a system is to succeed or fail. For example, calculating the probability of a Series component working involves multiplying the individual probabilities. Steps include:
  • Finding probabilities of individual components, like 0.9 in our exercise for each component working.
  • Using these to calculate series probabilities by multiplying: e.g., for subsystem 1-2, it's 0.9 x 0.9 = 0.81.
  • Determining parallel probabilities by understanding that the system will work if at least one path works, using complementary probabilities to find out potential failure rates.
Therefore, operations like these help us determine the overall reliability and guide improvements or changes to system design.

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Most popular questions from this chapter

Is ultrasound a reliable method for determining the gender of an unborn baby? Consider the following data on 1000 births, which are consistent with summary values that appeared in the online journal of Statistics Education ("New Approaches to Learning Probability in the First Statistics Course" 120011 ): \begin{tabular}{l|cc} & Ultrasound Predicted Female & Utrasound Predicted Male \\ \hline Actual Gender Is Female & 432 & 48 \\ Actual Gender is Male & 130 & 390 \\ \hline \end{tabular} Do you think that a prediction that a baby is male and a prediction that a baby is female are equally reliable? Explain, using the information in the table to calculate estimates of any probabilities that are relevant to your conclusion.

Jeanie is a bit forgetful, and if she doesn't make a "to do" list, the probability that she forgets something she is supposed to do is 1 . Tomorrow she intends to run three errands, and she fails to write them on her list. a. What is the probability that Jeanie forgets all three errands? What assumptions did you make to calculate this probability? b. What is the probability that Jeanie remembers at least one of the three errands? c. What is the probability that Jeanie remembers the first errand but not the second or third?

Five hundred first-year students at a state university were classified according to both high school grade point average (GPA) and whether they were on academic probation at the end of their first semester. The data are summarized in the accompanying table. \begin{tabular}{lcccc} & \multicolumn{3}{c} { High School GPA } & \\ \cline { 2 - 4 } Probation & 2.5 to & 3.0 to & 3.5 and & \\ & \(<3.0\) & \(<3.5\) & Above & Total \\ \hline Yes & 50 & 55 & 30 & 135 \\ No & 45 & 135 & 185 & 365 \\ Total & 95 & 190 & 215 & 500 \\ \hline \end{tabular} a. Construct a table of the estimated probabilities for each GPA-probation combination by dividing the number of students in each of the six cells of the table by \(500 .\) b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of 3.5 or above? d. Are the two outcomes selected student has a GPA of 3.5 or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between 2.5 and 3.0 who are on academic probation at the end of the first semester.

"N.Y. Lottery Numbers Come Up \(9-1-1\) on \(9 / 11^{*}\) was the headline of an article that appeared in the San Francisco Chronicle (September 13,2002\()\). More than 5600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. A professor at the University of Buffalo is quoted as saying, "Tm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up. People tend to read into these things. I'm sure that whatever numbers come up tonight, they will have some special meaning to someone, somewhere." The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. To select the winning sequence, one ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) would be the one selected on any particular day?

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, nine are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as much of the request 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 who wishes to be an independent driver 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?

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