Probability of 3 Events Calculator

Three independent events open a whole fan of questions — all three? none? at least one? exactly one? exactly two? — and each of those questions has an exact answer built from the same three numbers you supply. This calculator runs the complete fan: enter P(A), P(B) and P(C) as decimals between zero and one, and it returns every standard combination for three independent events, computed exactly by the complement and inclusion rules rather than by simulation, approximation or guesswork of any kind whatsoever.

The Fan of Questions, One Machinery

Three events generate eight elementary worlds (each event happens or not: 2³ = 8), and every standard question is a sum over some of those worlds. Independence prices each world as a product — ABC, AB¬C, and so on — and the art is purely in choosing which worlds answer the question asked. All three is the single world ABC. None is its mirror ¬A¬B¬C, computed by multiplying the complements. At least one is the seven remaining worlds, priced most cheaply as 1 − none, the complement rule doing the work of seven additions. Exactly one sums the three lone-success worlds; at least two is assembled by inclusion (pairwise joints, with the triple subtracted back once, per the calculator’s method). The fan looks like five formulas; it is one product table wearing five sums, which this page computes in the open. Mastering the table — rather than the five sums — is what separates someone who can price any combination of three events from someone who can only recite the combinations a textbook happened to name, and the generalisation to four, five or n events costs nothing once the table is understood properly.

Worked Example: Halves and Fractions

The defaults — P(A) = 0.5, P(B) = 0.4, P(C) = 0.3 — run the fan end to end. All three: 0.5 × 0.4 × 0.3 = 0.06. None: 0.5 × 0.6 × 0.7 = 0.21. At least one: 1 − 0.21 = 0.79. Exactly one: A alone 0.5 × 0.6 × 0.7 = 0.21, B alone 0.5 × 0.4 × 0.7 = 0.14, C alone 0.5 × 0.6 × 0.3 = 0.09 — totalling 0.44. At least two: pairwise joints 0.20 + 0.15 + 0.12 minus twice the triple (2 × 0.06) = 0.35. Audit the partition: none 0.21 + exactly one 0.44 + exactly two (0.35 − 0.06 = 0.29) + all three 0.06 = 1.00 — the eight worlds fully accounted, which is the calculator’s internal conscience and the check any hand computation should imitate.

Inclusion–Exclusion: The Union in Full

The union of three events is where the addition rule grows its full plumage: P(A∪B∪C) = [P(A) + P(B) + P(C)] − [P(A∩B) + P(A∩C) + P(B∩C)] + P(A∩B∩C). With the defaults: singles sum 1.20; pairs subtract 0.47; the triple adds back 0.06; the union = 0.79 — matching the complement route exactly, as it must. The alternating signs are the Venn diagram’s triple-overlap bookkeeping (the centre lens is added three times, subtracted three times, and must be restored once), and the pattern — singles, minus pairs, plus triples, minus quadruples — extends to any number of events as the principle of inclusion–exclusion. This calculator quietly uses the cheaper complement route for the union and the inclusion route for “at least two”, and shows both agreeing, because seeing the machinery agree from two directions is worth more than either answer.

When Independence Fails the Fan

Every figure on this page is rented from the independence assumption, and the rent is due whenever events share causes, compete for outcomes, or draw from a depleting pool. Three production machines fed by one power supply do not fail independently; three candidates for two jobs do not succeed independently; three cards drawn without replacement are the textbook’s standing warning. The repairs: replace products with conditional chains (the joint probability page’s machinery, extended), or model the shared cause explicitly and condition on it. What never repairs it is optimism — dependent worlds priced as independent understate conjunctions of positively-linked events (the shared-cause failures) and overstate them for competitors, in exactly the directions that embarrass risk assessments. Declare the assumption, as this page does on its face, or do not quote the fan.

A Neighbouring Case: Three Birthdays

The fan’s machinery illuminates a famous neighbour, provided the difference in model is respected. Do three random people include two who share a birthday? These are not three independent events of the calculator’s type (the “events” are entangled pairs), but the complement method still drives: the probability of no shared birthday is (364/365) × (363/365) ≈ 0.99180, so a match exists with probability ≈ 0.0082 — small for three people, on its way to the celebrated better-than-even at twenty-three. The doorway is worth one paragraph because it shows the complement rule’s universality: “at least one” questions of every species surrender to 1 − P(none), and the craft lies in pricing “none” correctly for the model at hand — products of complements here, falling factorials there.

When All Three Probabilities Are Equal

A special case with its own famous machinery: if the three events share one probability p (three identical machines, three flips of one biased coin), the fan becomes the binomial distribution with n = 3. The pattern: P(exactly k) = C(3,k) p^k (1−p)^(3−k). At p = 0.4: exactly two = 3 × 0.16 × 0.6 = 0.288; at least one = 1 − 0.6³ = 0.784; all three = 0.064. The calculator handles this case without knowing it (enter the same probability three times and compare against the binomial’s figures — they agree to the last decimal, a pleasant audit), while unequal probabilities, its native territory, need the general fan because no single p summarises them. Recognising when a problem secretly is binomial — identical, independent trials — is the pattern-matching skill the whole topic trains.

Reliability: The Fan at Work

Engineers meet this exact fan as system reliability. Three independent components with working probabilities 0.5, 0.4, 0.3 wired in parallel (the system works if any one works) deliver the at-least-one figure: 0.79 reliability from mediocre parts — redundancy’s whole promise. The same parts in series (all must work) deliver the all-three figure: a sobering 0.06. Identical components, identical probabilities, a thirteen-fold difference in system reliability decided purely by architecture — which is why reliability engineers draw the block diagram before reaching for any calculator, and why this page prints both ends of the fan in the same breath.

Related Tools

More from the Statistics Calculators hub: the Joint Probability Calculator for the two-event foundations, and the AND Probability Calculator for longer independent chains.

Frequently Asked Questions

What is the probability that at least one of three events happens?

One minus the probability that none do: 1 − (1−a)(1−b)(1−c) — 0.79 in the worked example. The complement route is always the cheapest path to “at least one”, for any number of events, which is why it is the first trick probability courses teach and the one this calculator headlines. The direct route (summing seven worlds or running inclusion–exclusion) agrees, at seven times the effort.

How do I get P(exactly two)?

Subtract the triple from “at least two”, or sum the three two-success worlds directly: AB¬C + A¬BC + ¬ABC = 0.14 + 0.09 + 0.06 = 0.29 in the example. The calculator prints “at least two” (0.35) and “all three” (0.06) explicitly so exactly-two is one subtraction away — and the partition audit (none + exactly one + exactly two + all three = 1) verifies all four at a glance.

Why does the union formula add the triple back?

Because the pairwise subtractions over-remove it. In the Venn picture, the centre region (all three events) is counted in all three singles (+3), removed in all three pairwise joints (−3), and must be restored once (+1) to be counted exactly once. Every term in inclusion–exclusion is that argument at one depth of overlap; the alternating signs are not a pattern to memorise but a debt being settled region by region.

Three events, all mutually exclusive: what changes?

Everything simplifies brutally: all joints are zero, exactly one = the plain sum a + b + c, at least one = the same sum, and none = 1 − (a+b+c). Exclusivity is a different universe from independence (the joint probability page dwells on their confusion), and this calculator’s fan presumes independence — entering exclusive events here would silently misprice them, the eternal argument for reading a calculator’s assumptions before its answers.

Can the probabilities be entered as percentages or fractions?

Convert first: 50% is 0.5 and 1/3 is 0.333 — the calculator works in the 0–1 probability scale throughout, and returns both decimals and percents. Fractions with small denominators are worth entering at full precision (0.3333 rather than 0.33) when exactness matters, since the fan multiplies three of them and rounding compounds politely but surely.

What if one event is certain or impossible?

The fan degenerates gracefully, and checking the degeneration is a fine audit: P(C) = 1 forces all three = P(A∩B), none = 0, at least one = 1; P(C) = 0 kills every world containing C and reduces the fan to the two-event case. The calculator handles both ends without complaint — boundary behaviour is where probability tools either earn their keep or reveal they were curve-fitted to the middle of the range and never honestly tested at its edges at all.