Calculations with context

Fault-tree calculator

Explore gate logic, minimal cut sets and conditional-probability assumptions in the fixed three-event model T = C OR (A AND B).

One shared cause. Two intrinsic failures.

Change three probabilities and see exactly how the shared supply and the two-channel branch contribute. This fixed learning model is T = C OR (A AND B). It does not construct or solve arbitrary fault trees.

P(T) = pC + (1 − pC) × pA|¬C × pB|¬C

Define the start demand first

Imagine a shipboard equipment cabinet with two parallel ventilation fans. Either fan alone provides the required airflow. Both receive a valid start command; they share one supply and the air path is clear. The top event T is failure to establish that airflow on one defined start demand. This is an invented teaching scenario, not field data or a claim about a real ship or product.

Event probabilities

All inputs are dimensionless probabilities per the same defined start demand. Enter 0–1, not percent, failure rate or events/year. For example, 0.002 means 0.2%. Use a decimal point or comma; no thousands separators.

P(C) · shared-supply unavailableC alone defeats both fan paths. Include this cause here only once.

P(A | not C) · intrinsic fan A failureFailure to start given adequate shared power. Exclude shared-supply loss from this input.

P(B | not C) · intrinsic fan B failureFailure to start given adequate shared power. Exclude shared-supply loss from this input.

For this teaching case, all three inputs refer to the same defined start demand, event conditions and exposure. The A and B values exclude C.

I assume that A and B are conditionally independent given that C has not occurred. I have considered remaining shared control, environment, maintenance and design dependencies.

Both acknowledgments are required before quantification. Checking them records an assumption; it does not verify independence or the applicability of any data.

Numerical limit: each value must be 0 or at least 10⁻¹⁰⁰, up to 1. This software precision floor prevents tiny comparison products from underflowing; it is not an engineering probability threshold.

Trace the probability

The method and worked example are available below. Enable JavaScript to change inputs.

The calculator processes inputs and creates CSV files locally in this page. It sends no calculator data and uses no persistent storage. A download starts only when you select Download result CSV.

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Fixed three-event logic

Gate labels describe Boolean logic. An AND gate alone never justifies multiplying marginal probabilities.

Check every logical state

Eight Boolean assignments for T = C OR (A AND B)
C	A	B	T
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	1

0 means false; 1 means true. These eight rows are logical assignments, not equally likely outcomes. No state probabilities are assigned. When C is true, the A/B assignments do not establish failure probabilities conditional on a functioning supply.

The two minimal cut sets

A cut set is sufficient to cause T. It is minimal when removing any member makes it insufficient.

{C}: shared-supply loss alone is sufficient.
{A, B}: both intrinsic fan failures are required; neither alone is sufficient when C is false.

“Minimal” describes set inclusion, not the lowest probability. C retains one identity even if a drawing repeats it: C OR C = C and C AND C = C. Its probability must never be squared as though the two appearances were independent events.

Why the probability formula is conditional

The Boolean structure comes first. Define C as shared-supply unavailability and define A and B as the intrinsic fan failure conditions when adequate power is present. If either fan cannot meet the required airflow alone, this tree is not the correct structure.

Partition the top event into C and the supply-available case: P(T) = P(C) + P(not C) × P(A AND B | not C). These two cases are mutually exclusive, so they can be added without double-counting.

In general, P(A AND B | not C) = P(A | not C) × P(B | A AND not C). Only under the stated conditional-independence assumption can the second factor be replaced by P(B | not C). Independence of A and B without conditioning is neither required nor established by this model.

The resulting expression is P(T) = pC + (1 − pC) × pA|notC × pB|notC. C is an explicit shared-supply dependency; this is not a fitted common-cause or beta-factor model. Additional shared causes must not be silently absorbed into supposedly independent fan data.

Do not put all-causes fan failure probabilities into A and B and then add the same shared-supply cause again as C. A rate per hour, a demand probability and an annual event frequency are different quantities. This calculator performs no rate-to-probability conversion.

Reproduce the guide’s worked example

The existing practical guide supplies invented per-start-demand values: pC = 0.002 and pA|notC = pB|notC = 0.01. The conditional independence and shared-basis assumptions are explicit.

Conditional joint failure: 0.01 × 0.01 = 0.0001.
Supply-available joint contribution: (1 − 0.002) × 0.0001 = 0.0000998.
Total: 0.002 + 0.0000998 = 0.0020998, or 0.20998% per defined start demand.
With C reduced to 0.0002: 0.0002 + 0.9998 × 0.01 × 0.01 = 0.00029998, or 0.029998%.
With C unchanged and both fan inputs halved: 0.002 + 0.998 × 0.005 × 0.005 = 0.00202495, or 0.202495%.

Here the hypothetical tenfold reduction in C has a larger effect than halving both intrinsic inputs. The alternatives are different changes, not equal-cost interventions. This sensitivity establishes neither feasible improvement nor a real-world engineering priority.

What this small tree leaves out

This is a static, non-repair start-demand model. It does not model running failure, repair, failure order, switching, duty/standby transfer, testing intervals or multiple operating modes. An AND gate does not mean “A happens before B.”
A common blocked air path, invalid start command, remaining shared environment or maintenance error could cause loss of airflow outside this selected tree. A low total cannot prove those causes are absent.
Input values are point assumptions with no confidence bounds or data fitting. Zero entered probability is a modeling boundary, not evidence that a failure is impossible. Values near numerical limits may lose displayed precision.
This tool has exactly three event identities and two specified gates. It cannot import, draw or solve an arbitrary network, reduce general cut sets, or decide whether duplicated basic events are independent.
No tolerability decision, SIL assignment, certification, equipment selection or permission to operate is produced. Real use requires suitable event evidence, dependency review and competent engineering assessment.

Method and further reading

Primary sources support method context; the fan data and comparisons are authored teaching assumptions. Sources checked 8 October 2026. Model version 1.0.0.

Read the method and its limits

Related project: AURA

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