Probability Calculator
The two-event algebra with the axioms enforced: the addition rule with the double-count removed, both conditional probabilities, an independence verdict from the multiplication rule, and refusals when P(A and B) leaves what the axioms allow.
Probability Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Two events only — inclusion–exclusion for three is out of scope
- Needs the overlap P(A∩B); it cannot be inferred from the marginals alone
In short: With P(A) = 0.3, P(B) = 0.5 and P(A∩B) = 0.15: P(A∪B) = 0.3 + 0.5 − 0.15 = 0.65 — the subtraction is the double-count leaving, since adding the two events counts their overlap twice. Both conditionals come back at the marginals (P(A|B) = 0.15/0.5 = 0.3, P(B|A) = 0.5), which is what independence looks like in numbers: P(A)P(B) = 0.15 = P(A∩B) exactly, the multiplication rule holding. The axioms police the inputs: P(A∩B) must sit between max(0, P(A)+P(B)−1) = 0 and min(P(A), P(B)) = 0.3 — claim 0.4 and the page refuses, because 0.1 of B would be happening outside B.
Formula
P(A∪B) = P(A)+P(B)−P(A∩B) · P(A|B) = P(A∩B)/P(B) · independent ↺ P(A∩B) = P(A)P(B)
The bounds max(0, P(A)+P(B)−1) ≤ P(A∩B) ≤ min(P(A), P(B)) are not conventions — they are what keeps a probability assignment coherent. Violations are refused with the violated wall named.
Worked Example
- Enter P(A), P(B) and the overlap P(A and B) — all between 0 and 1.
- The page checks the axiom walls first; violations name the bound they broke.
- Read the union, the two conditionals, and the independence verdict.
- Use the bounds card to see the allowable overlap window for your marginals.
Defaults: union 0.65, P(A|B) = 0.3, P(B|A) = 0.5, independent (0.3×0.5 = 0.15). P(A∩B) = 0 flips the verdict to mutually exclusive and NOT independent; 0.4 or 0.85 is refused on the walls.
Strengths & Limits Of This Model
Where this engine is strong
- Axiom walls enforced with the violated bound named
- Both conditionals plus the independence verdict from one panel
Where it stops
- No event trees or diagrams
- No Bayes inversion — the Bayes page owns the reverse direction
Practical Use Cases
Risk
two failure modes and their overlap
Teaching
addition rule, conditionals, independence
Data QC
is a claimed probability table even possible?
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Probability Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
P(A∪B) = 0.65 — where did the subtraction go?
Adding P(A) + P(B) counts the overlap twice: 0.3 + 0.5 counts the 0.15 they share two times. The addition rule subtracts it once: 0.3 + 0.5 − 0.15 = 0.65. The subtraction is not a convention; it is the double-count leaving.
Are A and B independent here?
Check the multiplication rule live: P(A)·P(B) = 0.3 × 0.5 = 0.15 = P(A∩B). That equality IS independence — each event tells you nothing about the other. The verdict card runs the comparison on whatever you type, with the two products shown.
What is P(A|B), in words?
The share of the B-world that is also A-world: P(A∩B)/P(B) = 0.15/0.5 = 0.3. Conditioning shrinks the universe to B and re-measures A inside it. With P(B) = 0 there is no world to stand on and the page refuses the division by name.
Mutually exclusive vs independent — same thing?
Opposites. Mutually exclusive: P(A∩B) = 0 — A happening FORBIDS B. Independent: P(A∩B) = P(A)P(B) — A happening tells you nothing about B. Two events with positive probabilities cannot be both: exclusion is information.
Why was P(A and B) = 0.4 refused?
The overlap cannot exceed the smaller event: P(A∩B) ≤ min(P(A), P(B)) = 0.3. At 0.4 you have claimed 0.1 of B happening outside B. The refusal names the wall — the numbers must satisfy the axioms before the algebra runs.
And the lower bound — where does it come from?
P(A∩B) ≥ max(0, P(A)+P(B)−1). With P(A) = 0.6 and P(B) = 0.7 the overlap must be at least 0.3, or the union would need probability 1.3 in a sample space that has only 1. The bounds card prints your standing against both walls.
Can I use this for three events?
The three-event inclusion–exclusion has eight terms and this page keeps to two — honestly. Chain it instead: collapse A∪B first, then feed that union’s probability and C’s, once you have their overlap from somewhere honest.
Where does Bayes fit in?
This page runs the union and intersection algebra FORWARD. The reverse direction — P(H|E) from P(E|H) — is Bayes’ theorem, and the Bayes page owns it with the base-rate tree. Same axioms, opposite conditioning direction; the two pages cross-link.