How Component Failures Affect a System
A device can contain several components, but whether the whole device works depends on how those components are arranged. If every component is essential, one failure can stop the system. If components provide alternatives, the system may keep working when one fails. Independence lets us calculate the probability of combinations of component outcomes, but first we must identify how the system is built.
In The Multiplication Rule for Independent Events and Independence of Three or More Events, you learned that the probability all of several mutually independent events occur is the product of their probabilities. Here, the events describe whether individual components work or fail. As emphasized in Common Errors with Independence and Unions, independence must be part of the model or justified by the process; it is not automatic just because components are different.
Two basic arrangements are called series and parallel. In a series system, every component must work for the system to work. In a parallel system, the system works if at least one component works. Those different success rules lead to different failure calculations.
Series Systems: One Failure Stops the System
Suppose a system has \(n\) components arranged in series. Let \(C_i\) be the event that component \(i\) works, and let \(R_i=P(C_i)\) be its reliability. The system works only if all \(n\) components work. If their working events are mutually independent, multiply their reliabilities to find the probability the system works. Then use the complement to find its failure probability.
The complement is useful because “the system fails” means at least one component fails. Instead of listing every possible combination with at least one failure, calculate the probability all components work and subtract from 1. This is an application of the complement idea used in Probability of At Least One in Independent Trials, with a key difference: component reliabilities do not have to be equal.
Worked Example: A Three-Sensor Series System
A hypothetical safety device uses three sensors in series. For the device to operate, all three sensors must work during a test. Their respective probabilities of working are \(0.98\), \(0.95\), and \(0.90\). Assume the sensors’ outcomes are mutually independent. Find the probability that the device fails during the test.
State. Let \(C_1,C_2,C_3\) be the events that sensors 1, 2, and 3 work. The target is the probability that the series system fails.
Plan. In a series arrangement, the device works only if all three sensors work. The sensors are stated to be mutually independent, so the probability all work is the product of their individual reliabilities. Subtract that probability from 1 to find the probability the device fails.
Do. First find the probability all three sensors work:
Therefore, the system’s failure probability is
As a check, the sensors’ failure probabilities are \(0.02\), \(0.05\), and \(0.10\). The chance that none fail is \((0.98)(0.95)(0.90)=0.8379\), so the chance that at least one fails is again \(1-0.8379=0.1621\).
Conclude. Under the stated independence model, the probability that the three-sensor series device fails during the test is \(0.1621\), or about \(16.21\%\). Even though each sensor individually has a high probability of working, the system requires all three to work.
Parallel Systems: Every Part Must Fail to Stop the System
In a parallel arrangement, the system works when one or more components work. Therefore, the system fails only when every component fails. If component \(i\) has reliability \(R_i\), its failure probability is \(1-R_i\). When component outcomes are mutually independent, multiply those failure probabilities to find the probability the entire parallel system fails.
Notice the contrast: the series formula multiplies probabilities of components working to find system reliability; the parallel formula multiplies probabilities of components failing to find system failure probability. Either arrangement can then be described using the complement, but it is important to identify which event the product represents.
Worked Example: Backup Pumps in Parallel
A hypothetical water system has three independent pumps arranged in parallel. The system can supply water as long as at least one pump works. The probabilities that pumps 1, 2, and 3 work during a power interruption are \(0.92\), \(0.90\), and \(0.95\). Find the system’s failure probability and reliability.
State. Let \(F_i\) be the event that pump \(i\) fails. The system fails if \(F_1,F_2,\) and \(F_3\) all occur.
Plan. The pumps’ outcomes are stated to be independent. Find each pump’s failure probability by subtracting its reliability from 1. Multiply those failure probabilities to find the probability all three fail. Then subtract from 1 to find system reliability.
Do. The individual failure probabilities are \(1-0.92=0.08\), \(1-0.90=0.10\), and \(1-0.95=0.05\). Thus,
The reliability is the complement:
The result is consistent with the system’s design: failure requires all three pumps to fail, not merely one. For another arithmetic check, the product of failure probabilities is \(0.0004\), so the probability of at least one working pump is \(1-0.0004=0.9996\).
Conclude. In this hypothetical model, the parallel pump system has a failure probability of \(0.0004\), or \(0.04\%\), and a reliability of \(0.9996\), or \(99.96\%\), during the power interruption.
Systems with More Than One Arrangement
Some systems combine series and parallel structures. A useful strategy is to divide the system into paths, determine what must happen for each path to work, and then determine what combination of path outcomes makes the whole system work. For two parallel paths, the system fails if both paths fail. If the paths use separate, mutually independent components, their success or failure events are independent as well.
For example, if each path contains two components in series, a path works only when both of its components work. Find each path’s reliability by multiplying its components’ reliabilities. Then treat the paths as parallel: the system fails when both paths fail. This approach keeps the component-level and system-level logic separate.
Worked Example: Two Parallel Paths with Series Components
A hypothetical monitoring system has two separate paths. Path A contains two components in series, with probabilities of working \(0.90\) and \(0.80\). Path B also contains two components in series, with probabilities of working \(0.95\) and \(0.90\). The whole system works if either path works. Assume all four component outcomes are mutually independent. Find the probability the whole system fails.
State. Let \(A\) be the event that Path A works and \(B\) the event that Path B works. Because each path is a series arrangement, both of its components must work. Because the paths are in parallel, the whole system fails when both paths fail.
Plan. First calculate each path’s reliability by multiplying the reliabilities of the two components in that path. Then find each path’s failure probability. The paths use separate components from a mutually independent set, so their outcomes are independent; multiply the two path-failure probabilities to get the system-failure probability.
Do. Path A’s reliability is
Path B’s reliability is
The path-failure probabilities are \(1-0.72=0.28\) and \(1-0.855=0.145\). Both paths must fail for the system to fail, so
As a check, the probability the system works is \(1-0.0406=0.9594\). This is also the chance that Path A works, Path B works, or both work: \(0.72+0.855-(0.72)(0.855)=1.575-0.6156=0.9594\).
Conclude. Under the stated model, the probability that both paths fail and the monitoring system stops working is \(0.0406\), or \(4.06\%\). Its reliability is \(0.9594\), or \(95.94\%\).
Common Mistakes and AP Exam Tips
- Multiplying the wrong probabilities. In a series system, multiply component working probabilities to find the probability all work. In a parallel system, multiply component failure probabilities to find the probability all fail. State what the product represents before using it.
- Confusing “at least one works” with “all work.” A series system needs all components to work. A parallel system needs at least one to work. Read the system description before choosing the event to calculate.
- Forgetting the complement. A product may give the probability of all components working or all components failing, not the probability requested. If the target is the opposite event, subtract the product from 1.
- Assuming independence without support. Components can share a power supply, environment, or other cause of failure. Use the product rule only when independence is specified or justified by the system model. For several components, the model must support mutual independence for the all-components product.
- Rounding too early. Keep the given reliabilities through the multiplication and round the final probability. A small error in a component probability can affect the system result, especially across several components.
For full-credit communication, name the arrangement, define the event being calculated, justify the independence assumption, and interpret the result in context. For example: “The system is parallel, so it fails only if all three independent pumps fail. Multiplying their failure probabilities gives \(0.0004\); therefore, the probability the system fails during the interruption is \(0.0004\).” Avoid writing only a product without explaining why that product matches the system’s failure rule.
Key Takeaway
System reliability depends both on component reliabilities and on the arrangement of the components. Series systems fail when at least one required component fails, while parallel systems fail only when all their components fail. Independence makes the relevant products available; the system’s structure determines which probabilities to multiply.
Check Your Understanding
For each question, identify whether the system is series or parallel and show which event the product represents.
- Two independent components in series have reliabilities \(0.96\) and \(0.85\). Find the system’s reliability and failure probability.
- Two independent components in parallel have failure probabilities \(0.04\) and \(0.12\). Find the system’s failure probability and reliability.
- A series system has three mutually independent components with reliabilities \(0.80\), \(0.90\), and \(0.95\). Explain why multiplying these values does not directly give the failure probability.
- A parallel system has three independent components, each with reliability \(0.90\). Find the system’s failure probability. Show the component failure probability first.
- A system has two separate paths in parallel, each path consisting of two independent series components. What probability must be found first for each path, and what must happen for the whole system to fail?