Probability Calculator

Compute single-event, union, intersection, complement and at-least-once probabilities for independent, mutually exclusive or overlapping events — including how many tries a 95% chance needs.

Updated August 2026 Math & Statistics

Enter two probabilities

How the events relate
Probability of A or B
A and B together
Neither happens
A at least once
Tries for a 95% chance

About precision. Results are computed in double-precision floating point, which is exact for whole numbers up to about sixteen digits and very slightly approximate beyond that. Figures are rounded for display, so a long chain of calculations can differ from a hand-worked answer in the last decimal place. Where an exact fraction or radical exists, this page shows it alongside the decimal.

How to Use the Probability Calculator

Most probability questions in ordinary life are one of four: how likely is this, how likely are both, how likely is either, and how likely is it to happen at least once if I keep trying. This calculator answers all four, and the last one is the one that surprises people.

  1. Enter the two probabilities as percentages. A 35% chance goes in as 35, not 0.35. The calculator converts internally.
  2. Choose how the events relate. Independent means one does not affect the other — two separate coin flips. Mutually exclusive means they cannot both happen. Known overlap is for when you have measured how often both occur.
  3. Set the number of tries. This drives the at-least-once figure, which is the calculation behind everything from lottery odds to whether a rare software bug will show up in testing.
  4. Read the union and intersection. The union — A or B — subtracts the overlap once, because adding two probabilities counts the both case twice.
  5. Check the tries-for-95% tile. It answers the practical version of the question: how many attempts before you can be reasonably confident of seeing the event at least once.

Independence is an assumption, not a fact, and it is the assumption most often wrong. Two mortgage defaults in the same town are not independent events, which is a large part of why 2008 happened.

Probability Rules

Five rules, and the at-least-once one is the least intuitive.

Complement: P(not A) = 1 − P(A)Independent intersection: P(A and B) = P(A) × P(B)Union: P(A or B) = P(A) + P(B) − P(A and B)Mutually exclusive: P(A and B) = 0, so P(A or B) = P(A) + P(B)At least once in n: 1 − (1 − P(A))nConditional: P(A given B) = P(A and B) ÷ P(B)The at-least-once rule works by inversion: instead of adding up every way the event could happen, calculate the single way it could not — never, in any of the n tries — and subtract that from one.
What each symbol means
SymbolMeaningUnitTypical range
P(A)Probability of the first event%0 – 100
P(B)Probability of the second event%0 – 100
nNumber of independent triescount1 – 100,000
UnionEither event happens%0 – 100
IntersectionBoth happen%0 – 100

Subtracting the intersection from the union is the step people forget. Adding two probabilities counts every outcome where both occur twice — once in each term — so it has to come off once to get back to the truth. Where the events cannot both happen, the intersection is zero and the subtraction has nothing to do.

Example

Two independent events at 35% and 20%

  1. Both happen: 0.35 × 0.20 = 7.00%.
  2. Either happens: 0.35 + 0.20 − 0.07 = 48.00%. Note it is not 55% — the overlap would be counted twice.
  3. Neither happens: 1 − 0.48 = 52.00%, which is also 0.65 × 0.80.
  4. A does not happen: 1 − 0.35 = 65.00%.
  5. A never happens in eight tries: 0.658 = 3.19%.
  6. A happens at least once in eight tries: 1 − 0.0319 = 96.81%.

Why at-least-once climbs so fast

A single 35% chance feels unlikely. Eight tries at 35% is 96.81% likely to produce at least one success — near enough a certainty. That gap between intuition and arithmetic is behind a great many surprises: rare events become close to inevitable given enough opportunities, which is why a one-in-a-million manufacturing fault matters when you make ten million units.

How many tries for 95% confidence?

Solving 1 − 0.65n ≥ 0.95 gives n ≥ ln(0.05) ÷ ln(0.65) = 6.95, so 7 tries. That is the practical form of the question, and it is the same arithmetic that decides how many test runs are needed before you can say a one-in-twenty bug probably is not there.

Why Rare Events Are Not Rare Enough

How the at-least-once probability builds with repeated tries at 35%.

The chance of seeing A at least once, and of never seeing it
TriesAt least onceNeverNote
135.00%65.00%A single attempt
372.54%27.46%
588.40%11.60%
896.81%3.19%Your figure
1299.43%0.57%
2099.98%0.02%Effectively certain

The right-hand column is the one doing the work. Never seeing the event requires it to fail every single time, and each additional try multiplies that requirement by 0.65 — so the chance of a clean sweep of failures collapses geometrically while the chance of at least one success climbs towards certainty.

The same arithmetic run backwards explains why rare things happen constantly. An event with a one-in-a-million chance per attempt is essentially impossible once. Given a million attempts, the chance of it happening at least once is 63.2% — which is why extremely unlikely coincidences are reported somewhere every day, and why a system handling millions of transactions will meet its million-to-one edge case.

Where this calculator stops is conditional probability in any depth. The dependent mode takes a joint probability you supply and derives the conditional from it, but it cannot help you estimate that joint figure — and that estimation is where most real probability errors live, because human intuition about how much one event tells you about another is famously poor.

Five Ways Probability Intuition Goes Wrong

Five ways probability intuition goes wrong.

Assuming independence. Two events are independent only if knowing one tells you nothing about the other. Rain today and rain tomorrow are not independent; two shipments from the same factory failing are not independent; and multiplying probabilities when they are correlated understates joint risk, sometimes catastrophically.

The gambler's fallacy. A fair coin that has landed heads six times is still exactly 50% to land heads next. The coin has no memory. What is unlikely is seven heads in advance, at 0.78% — but six of them have already happened, and past events do not change future ones.

Confusing P(A given B) with P(B given A). Almost everyone with a rare disease tests positive; very few people who test positive have the disease. These are different questions and the answers can differ by orders of magnitude when the underlying condition is rare.

Ignoring the base rate. A test that is 99% accurate for a condition affecting one person in ten thousand produces far more false positives than true ones, simply because there are so many more healthy people to be wrong about. This is the single most consequential probability error in medicine and in security screening alike.

Forgetting how many chances there were. The odds of any one person winning a lottery are minuscule; the odds of somebody winning are high. Both are correct and they answer different questions, which is what the at-least-once calculation above makes concrete.

The practical defence against all five is to state the reference class out loud. "A 1% chance" of what, out of how many attempts, among which population, over what period? Almost every misleading probability statement in circulation is missing at least one of those four, and supplying them usually resolves the argument without any arithmetic at all.

It also makes the numbers checkable. A stated probability with its reference class attached can be tested against the record; one without is unfalsifiable, which is a poor property for a number that is meant to inform a decision.

For the descriptive side of the same subject — summarising data rather than predicting it — the standard deviation calculator and the average calculator cover spread and centre respectively.

Frequently Asked Questions

Multiply them, if they are independent. A 35% chance and a 20% chance give 0.35 × 0.20 = 7%.

Add them and subtract the overlap: 0.35 + 0.20 − 0.07 = 48%. Adding alone would give 55% and count the both-case twice.

The chance of seeing the event on any of several tries. It is easiest calculated by inversion: one minus the chance of never seeing it. At 35% over eight tries that is 96.81%.

Divide the natural log of 0.05 by the natural log of one minus the probability. At 35% that is 6.95, so seven tries.

The two events cannot both happen, so the intersection is zero and the union is simply the two added. Rolling a three and rolling a five on one die are mutually exclusive.

The chance of A given that B happened: P(A and B) ÷ P(B). With a 7% joint probability and B at 20%, P(A given B) is 35% — the same as P(A), which is what independence means.

No, and assuming so is the most common error in the subject. Independence means knowing one tells you nothing about the other, which is rarely true of real-world events.

No. A fair coin has no memory, so the next flip is 50% regardless. Seven heads in advance is unlikely at 0.78%; six that have already happened change nothing.

Because of how many chances there are. A one-in-a-million event has a 63.2% chance of occurring at least once in a million attempts, which is why remarkable coincidences are reported constantly.

Ignoring how common something is before applying a test. A 99% accurate test for a one-in-ten-thousand condition produces far more false positives than true ones, because there are so many more healthy people.