Source: Principles of Statistics I, Texas A&M University
Tags: system reliability, series system, parallel system, probability of failure, probability of success, component reliability, complex systems, redundancy
System reliability problems ask you to find the overall probability that a multi-component system works (or fails). The two building blocks are series (all must work) and parallel (only one needs to work). Complex systems are solved by identifying series and parallel sub-sections, simplifying from the inside out.
Series system
A system where every component must function for the system to work. If any single component fails, the whole system fails. P(Success) = product of all individual component reliabilities.
Parallel system
A system where at least one component must function for the system to work. The system fails only if every component fails simultaneously.
Component reliability
The probability that an individual component works, typically given as a decimal between 0 and 1.
System reliability, P(S)
The probability that the overall system functions correctly.
System failure probability, P(F)
P(F) = 1 − P(S). Often easier to calculate directly for parallel systems.
Redundancy
Adding parallel paths so the system can survive individual component failures. More parallel paths means higher reliability, all else equal.
Components are arranged in a single path from input to output.
The system works only if A works AND B works AND C works (and so on).
P(S_series) = P(A) × P(B) × P(C) × …
Because you multiply probabilities that are each less than 1, adding more components in series always lowers system reliability.
Example: Four components in series with reliabilities 0.9, 0.8, 0.7, 0.6.
P(S) = 0.9 × 0.8 × 0.7 × 0.6 = 0.3024
The series chain is only as strong as its weakest link. Even though each component is fairly reliable on its own, the overall system succeeds less than a third of the time.
Components provide alternative paths from input to output.
The system fails only if every parallel component fails at the same time.
Calculate failure first, then convert:
P(F_parallel) = P(F_A) × P(F_B) × … = (1 − P(A)) × (1 − P(B)) × …
P(S_parallel) = 1 − P(F_parallel)
Example: Two components in parallel with reliabilities 0.7 and 0.6.
P(F) = (1 − 0.7) × (1 − 0.6) = 0.3 × 0.4 = 0.12
P(S) = 1 − 0.12 = 0.88
Even though neither component is brilliant on its own, the parallel arrangement gives you 88% reliability.
If there is more than one path from the left side to the right side of the system, the overall structure is a parallel.
If there is only one path, it is a series.
Sub-sections within a larger system can be series or parallel. You simplify from the inside out.
The worked example from the source has nine components (A through I) arranged as:
Top leg: A → B → C → D in series
Bottom leg: a parallel pair (E, F), then G in series, then another parallel pair (H, I)
Overall: the top leg and the bottom leg are in parallel with each other
Step 1: Solve each small parallel sub-system.
Left parallel (E and F): P(F) = (1 − 0.7)(1 − 0.6) = 0.12, so P(S) = 0.88
Right parallel (H and I): P(F) = (1 − 0.5)(1 − 0.4) = 0.30, so P(S) = 0.70
Step 2: Solve each leg as a series of its simplified blocks.
Top leg: P(S_top) = 0.9 × 0.8 × 0.7 × 0.6 = 0.3024
Bottom leg: P(S_bottom) = 0.88 × 0.9 × 0.70 = 0.5544
Step 3: Convert each leg to failure probability.
P(F_top) = 1 − 0.3024 = 0.6976
P(F_bottom) = 1 − 0.5544 = 0.4456
Step 4: The two legs are in parallel, so the system fails only if both legs fail.
P(F_overall) = 0.6976 × 0.4456 = 0.31085
P(S_overall) = 1 − 0.31085 ≈ 0.689, or about 68.9%
Redraw the system diagram if it helps. Label every component.
Identify the innermost series or parallel groups first.
Replace each group with a single equivalent component whose reliability you have just calculated.
Repeat until the entire system is reduced to one number.
For parallel sections, always calculate failure first (multiply failure probabilities), then convert to success.
For series sections, multiply success probabilities directly.
Series reliability: P(S) = P(C₁) × P(C₂) × … × P(Cₙ)
Parallel failure: P(F) = (1 − P(C₁)) × (1 − P(C₂)) × … × (1 − P(Cₙ))
Parallel reliability: P(S) = 1 − P(F) = 1 − [(1 − P(C₁)) × (1 − P(C₂)) × … × (1 − P(Cₙ))]
Complement (converting between success and failure): P(F) = 1 − P(S) and P(S) = 1 − P(F)
⚠️ The most common error is multiplying success probabilities for a parallel system. Parallel systems should be solved via failure: multiply the failure probabilities of all branches, then take the complement.
⚠️ When a system has nested structures (parallel within series within parallel), work from the inside out. Trying to solve the whole thing in one step is where mistakes happen.
⚠️ Series reliability can only decrease as you add components. Parallel reliability can only increase as you add components. If your answer goes the wrong direction, you have likely mixed up the two.
⚠️ On an exam, always double-check that your final reliability is between 0 and 1 and is plausible given the component values. A parallel system should be more reliable than its best single component. A series system should be less reliable than its weakest component.
Q: Two components are in parallel with reliabilities 0.9 and 0.8. What is the system reliability?
A: P(F) = (0.1)(0.2) = 0.02. P(S) = 1 − 0.02 = 0.98.
Q: Three components are in series with reliabilities 0.95, 0.90, and 0.85. What is the system reliability?
A: P(S) = 0.95 × 0.90 × 0.85 = 0.72675, or about 72.7%.
Q: A student calculates the reliability of a parallel system as 0.9 × 0.8 = 0.72. What did they do wrong?
A: They multiplied success probabilities, which is the formula for a series system. For a parallel system, multiply the failure probabilities (0.1 × 0.2 = 0.02) and then take the complement (1 − 0.02 = 0.98).
Q: In the complex system example, why is the overall structure classified as parallel even though the bottom leg contains series components?
A: Because there are two distinct paths from input to output (top leg and bottom leg). The existence of multiple paths makes the overall structure parallel, regardless of how the individual legs are configured internally.
Q: Adding a fourth component in parallel to a three-component parallel system will do what to overall reliability?
A: Increase it. Each additional parallel path provides another way for the system to succeed, reducing the probability that all paths fail simultaneously.
system reliability, series system, parallel system, series-parallel, complex system reliability, component reliability, redundancy, probability of failure, probability of success, reliability engineering, complement rule, nested systems, STAT 301, Principles of Statistics I, Texas A&M