A fault tree separates failure logic from probability assumptions
Derive minimal cut sets, retain repeated-event identity, and quantify overlapping failure combinations without multiplying dependent events blindly.
On this page
The question for today
What you’ll understand
- Translate a concrete top failure into Boolean event logic.
- Distinguish minimal cut sets from a smallest-cardinality cut set.
- Calculate an overlapping union with explicit independence assumptions.
Before estimating how often a service fails, define what counts as failure and which combinations can cause it. A fault tree organizes that logic from a top event down to named basic events. The logic can be correct even when the probability model is wrong. Keep these two tasks separate: first preserve causal identities and sufficient combinations, then ask which observed or assumed probabilities justify a numerical estimate.
Name events and the operating conditions
A top event should describe an observable failure under a stated mission or service condition, such as neither of two routes serving an eligible request. An OR gate means any input event suffices; an AND gate means their joint occurrence is required under the modeled conditions. These are logical relationships, not independence claims. A repeated basic event represents the same event wherever it appears. Duplicating its drawing does not create another statistically independent power supply, credential, or deployment action. Boundary conditions and omitted mechanisms should be documented beside the tree.
A cut set is a collection of basic failures sufficient for the top failure. It is minimal when removing any member makes that collection no longer sufficient; minimal does not mean it has fewer members than every other cut set. A tree can have both a one-event minimal cut and a three-event minimal cut. Listing them exposes single points and combinations worth investigating. The list is a consequence of the chosen logic, not proof the model includes every real failure mechanism or every dynamic sequence.
Overlapping cut sets cannot simply be added
The same basic-event combination can satisfy several cut sets. Adding their probabilities then counts that outcome more than once. Inclusion–exclusion corrects overlap; a rare-event sum can be a useful approximation or upper bound under explicit conditions, but it is not an exact result merely because each leaf has a small probability. Gate input events can share leaves even when the basic leaves are independent. Independence of A, B, and C therefore does not make the composite events A-and-B and A-and-C independent.
Quantification also needs compatible horizons and meanings. A one-hour component failure probability is not interchangeable with a per-request unavailability or a rate per hour. A static tree describes combinations within its chosen event definitions; order-sensitive recovery and standby behavior can need a dynamic model. Parameter uncertainty and missing common causes can dominate the approximation error from overlap. Use the tree to identify which assumptions matter, then collect evidence for them instead of reporting many decimal places from unsupported leaf numbers.
Challenge the model with a concrete failure trace
Walk through an incident and ask which basic events were true, which gate first became sufficient, and which mechanism the tree omitted. Record successful failover as well as failure: a redundant path may exist but fail to activate or lack capacity. A change that removes one cut set can create another dependency through its control plane. Review the logic when architecture or operating mode changes. A fault tree is useful because its assumptions can be inspected, not because naming gates turns uncertain architecture into a certified reliability figure.
The probability lab handles one special threshold case. Reset, set BOTH p=0.1 and n=3, interpret a success as a component failure, and predict P(at least two failures)=0.028 before Play or Step. Reset and re-set BOTH p=0.2 and n=3; predict 0.104. The exact binomial bars assume identical independent trials and do not build a fault tree, track shared causes, or sample incidents. They illustrate overlapping two-failure combinations in a symmetric case; use the worked event algebra for the different leaf probabilities below.
Work it through
Subtract the shared triple-failure outcome
For F=(A and B) or (A and C), assume mutually independent basic failures with P(A)=0.01, P(B)=0.02, and P(C)=0.03 over the same fictional mission.
The two cut-set probabilities are 0.01×0.02=0.0002 and 0.01×0.03=0.0003. Their naive sum is 0.0005.
Both cut sets occur when A, B, and C all occur, with probability 0.01×0.02×0.03=0.000006 under the stated mutual independence.
Subtract that overlap once: P(F)=0.0005−0.000006=0.000494, or 0.0494 percent per defined mission.
The exact probability is 0.000494 under this model. The repeated A makes the cut-set events overlap even though the basic events are independent.
The living notebook
Explore the model
Interactive lab · probability
Overlapping combinations in a symmetric threshold
For EACH run Reset, set BOTH p and n, predict, then Play or Step. Interpret success as failure: n=3 with p=0.1 gives P(K≥2)=0.028; p=0.2 gives 0.104. Exact independent binomial counts do not construct arbitrary fault trees, infer common causes, or sample failures.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA threshold union needs overlap-aware counting; independent leaves do not make every composite failure event independent.
Pause & reason
Check your understanding
Your turn
Try it yourself
A top failure is A or (B and C). List its minimal cut sets. If A is also drawn inside another redundant branch, should that drawing be treated as an independent copy of A?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Test whether each candidate remains sufficient after removing a member, and preserve event identity across the diagram.
Compare with the solution
The minimal cut sets are {A} and {B,C}. Their different cardinalities do not invalidate either minimal set. Another drawing of the same A remains the same basic event, so its probability and overlap must not be modeled as an independent new component.
Keep the idea
Take it with you
- Fault-tree logic and probability dependence require separate reasoning.
- Minimal cut sets expose sufficient combinations without implying equal size.
- Repeated events and overlapping cut sets need identity-preserving quantification.
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