How do I model "given the test was positive"?

Use Bayes via conditional: P(disease|positive) = P(positive|disease)·P(disease) / P(positive). Set A = disease, B = positive test; enter P(A) (prevalence) and P(A∩B) = P(positive|disease)·P(disease).

Why is P(A∩B) capped at min(P(A), P(B))?

Probability axiom: an event cannot be more likely than its constituents. P(A∩B) ≤ P(A) (since A∩B is a subset of A), so the joint can't exceed either marginal. The tool clamps and warns if you enter an inconsistent value.

Kolmogorov AN (1933), "Foundations of the Theory of Probability" — axiomatic definition. Modern textbooks: Ross SM (2019) "A First Course in Probability" 10th ed; DeGroot & Schervish (2011) "Probability and Statistics" 4th ed.

How do I model "given the test was positive"?

Use Bayes via conditional: P(disease|positive) = P(positive|disease)·P(disease) / P(positive). Set A = disease, B = positive test; enter P(A) (prevalence) and P(A∩B) = P(positive|disease)·P(disease).

Why is P(A∩B) capped at min(P(A), P(B))?

Probability axiom: an event cannot be more likely than its constituents. P(A∩B) ≤ P(A) (since A∩B is a subset of A), so the joint can't exceed either marginal. The tool clamps and warns if you enter an inconsistent value.

Kolmogorov AN (1933), "Foundations of the Theory of Probability" — axiomatic definition. Modern textbooks: Ross SM (2019) "A First Course in Probability" 10th ed; DeGroot & Schervish (2011) "Probability and Statistics" 4th ed.

Probability Calculator (P(A), P(A∩B), P(A|B))

Basic event probabilities + union + intersection + conditional + Bayes' theorem from two-event inputs.

Result

P(A∩B) · P(A∪B)

12.000% · 58.000%

P(A|B) = 30.000%; P(B|A) = 40.000%.

P(A)30.000%
P(B)40.000%
P(¬A)70.000%
P(¬B)60.000%
P(A∩B) — intersection (joint)12.0000%
P(A∪B) — union58.0000%
P(A|B) — A given B (conditional)Bayes-like: P(A|B) = P(A∩B) / P(B).30.0000%
P(B|A) — B given AP(B|A) = P(A∩B) / P(A).40.0000%
P(A∩¬B)18.0000%
P(¬A∩B)28.0000%
P(¬A∩¬B)42.0000%
Independence checkAsserted independent ⇒ P(A∩B) = P(A)·P(B).

Step-by-step

Independence assumed: P(A∩B) = P(A)·P(B) = 30.00% · 40.00% = 12.0000%.
P(A∪B) = P(A) + P(B) − P(A∩B) = 30.00% + 40.00% − 12.0000% = 58.0000%.
P(A|B) = P(A∩B) / P(B) = 12.0000% / 40.00% = 30.0000%.

How to use this calculator

Enter P(A) and P(B) as percentages.
Pick the relationship: independent (default for fair coins, dice, separate trials), mutex (cannot both happen), or custom joint (enter P(A∩B) directly for known correlation).
The tool shows all 2×2 contingency cells plus both conditional probabilities (A|B and B|A).

About this calculator

Three classical probability quantities from two-event inputs. P(A∩B) is the joint probability — both events occur. P(A∪B) is the union — at least one occurs, via the inclusion-exclusion principle P(A∪B) = P(A) + P(B) − P(A∩B). P(A|B) is the conditional probability of A given B, defined as P(A∩B)/P(B) (the proportion of A within B). Independence is a special case where P(A∩B) = P(A)·P(B), equivalent to P(A|B) = P(A) — B tells you nothing about A. Mutually exclusive events have P(A∩B) = 0 (they cannot co-occur). The tool also surfaces the 2×2 contingency-table cells: P(A∩B), P(A∩¬B), P(¬A∩B), P(¬A∩¬B), which always sum to 1. Bayes' theorem is implicit in the conditional outputs: P(B|A) = P(A∩B)/P(A) — useful when the conditional direction matters (e.g. updating from prior to posterior).

Frequently asked

Independent: B happening doesn't change P(A) (and vice versa). Mutually exclusive: B happening MAKES P(A) = 0 (they can't both happen). These are DIFFERENT concepts often confused — non-trivial events that are mutually exclusive are NEVER independent.