Calculate combined probability of three events using inclusion-exclusion. Supports independent and dependent events with Venn diagram regions, conditional probabilities, and repeated trials.
The probability of three events calculator computes all possible combinations and intersections of three events A, B, and C. Using the inclusion-exclusion principle, it finds P(A∪B∪C), the probability of exactly k events occurring, conditional probabilities, and all eight Venn diagram region probabilities.
For independent events, intersections are computed automatically (P(A∩B) = P(A)×P(B)). For dependent events, enter each pairwise and triple intersection manually to model correlated events like exam scores, weather patterns, or medical conditions.
The calculator also shows what happens over multiple trials, pairwise independence tests comparing actual vs expected intersections, and a complete breakdown of every possible outcome. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units. Cross-check this output using a known reference case. Use the example pattern when troubleshooting unexpected results. Validate that outputs match your chosen standards. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units.
Many real-world scenarios involve three or more events: passing three approval stages, detecting any of three defects, or weather affecting three crop types. This calculator provides the complete probability landscape for three events — something that's error-prone to calculate by hand due to the many overlapping regions.
Ideal for students learning probability theory, engineers modeling multi-component systems, and analysts evaluating combined risk scenarios.
P(A∪B∪C) = P(A) + P(B) + P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C). P(exactly 1) = P(A) + P(B) + P(C) − 2P(A∩B) − 2P(A∩C) − 2P(B∩C) + 3P(A∩B∩C).
Result: P(A∪B∪C) = 68.5%, P(A∩B∩C) = 3%, P(none) = 31.5%
With independent events, P(A∩B) = 12%, P(A∩C) = 10%, P(B∩C) = 7.5%, P(A∩B∩C) = 3%. Applying inclusion-exclusion: 40 + 30 + 25 − 12 − 10 − 7.5 + 3 = 68.5%. Exactly one event occurs 47.5% of the time.
While the two-event formula P(A∪B) = P(A) + P(B) − P(A∩B) is well-known, the three-event version adds complexity. The alternating sign pattern (add, subtract, add) generalizes: for n events, include/exclude subsets of increasing size with alternating signs. Understanding this pattern is fundamental to combinatorial probability.
Three events can be pairwise independent yet not mutually independent. If P(A∩B) = P(A)P(B), P(A∩C) = P(A)P(C), and P(B∩C) = P(B)P(C), but P(A∩B∩C) ≠ P(A)P(B)P(C), the events are pairwise but not mutually independent. The pairwise comparisons and triple comparison in this calculator help diagnose such subtle dependencies.
In system reliability, components often have three failure modes. The probability that at least one failure occurs, or that all three occur simultaneously, directly uses the three-event framework. Series systems fail when any component fails (union), while parallel systems fail only when all fail (intersection).
It prevents double-counting when computing the union of events. For three events: add individual probabilities, subtract pairwise intersections, then add back the triple intersection. Without this correction, overlapping regions would be counted multiple times.
Events are independent if P(A∩B) = P(A)×P(B) for every pair. In practice, use domain knowledge: coin flips are independent, but exam scores in related subjects are not. The pairwise comparison table helps check independence numerically.
The inclusion-exclusion principle extends to any number of events, but the formula grows exponentially (2^n − 1 terms). For four events, you'd need 15 terms. This calculator is designed specifically for three events.
It means one and only one of A, B, C occurs — the other two do not. This is the exclusive-or (XOR) for three events. It equals P(A only) + P(B only) + P(C only).
Negative values mean the input probabilities are inconsistent — the intersections violate probability axioms. For example, P(A∩B) cannot exceed P(A) or P(B). Adjust your dependent-event inputs to ensure mathematical consistency.
Each trial independently draws the three events. The chance of "all three" happening at least once in n trials is 1 − (1 − P(A∩B∩C))^n. This assumes trials are independent of each other.