How this instrument works
The chance that at least one of two events happens is P(A) plus P(B), minus the chance both happen together: P(A or B) = P(A) + P(B) - P(A and B). The subtraction matters — without it, any outcome where both A and B occur would get counted twice, once inside P(A) and again inside P(B), inflating the answer.
When the two events can never happen together — rolling a 2 and a 5 on the same single die roll, say — P(A and B) is 0, and the formula collapses to plain addition: P(A) + P(B). Statisticians call that case mutually exclusive, and it's the simpler special case of the general rule above.
This instrument needs P(A and B) as an input, not something it derives for you. If A and B are independent, this site's and-probability calculator multiplies P(A) x P(B) to find that overlap first; if they're dependent, the conditional-probability calculator gets you there instead.
- Enter the first event's chance into P(A) (%).
- Enter the second event's chance into P(B) (%).
- Enter the chance both happen together into P(A and B) (%) — 0 if mutually exclusive — leave it at 0 when the two events can never occur at the same time.
- Read P(A or B) (%) beneath them — the chance that at least one of the two events happens.
- Don't know P(A and B) yet? If the events are independent, this site's and-probability calculator multiplies P(A) x P(B) to find it first.
Worked example — at least one head in two coin flips
Set P(A) (%) to 50, P(B) (%) to 50, and P(A and B) (%) — 0 if mutually exclusive to 25 — the chance of heads on flip one, heads on flip two, and both landing heads together. P(A or B) (%) reads 75%, since 50 + 50 - 25 = 75. Direct enumeration confirms it: of the four equally likely outcomes HH, HT, TH, TT, three of them — HH, HT, TH — contain at least one head.
Set P(A and B) (%) — 0 if mutually exclusive back to 0 with two mutually exclusive events instead, like a 30% chance and a 20% chance that can never both occur, and the formula simplifies to plain addition: 30 + 20 - 0 = 50%. No overlap means nothing to subtract.
Questions
Why subtract P(A and B) instead of just adding P(A) and P(B)?
Because any outcome where both A and B happen is already counted once inside P(A) and again inside P(B). Adding them without correction double-counts that overlap, which can even push the total above 100% for events that overlap heavily. Subtracting P(A and B) removes the duplicate count exactly once, leaving the true probability that at least one event occurs.
What does 'mutually exclusive' mean, and when is P(A and B) zero?
Mutually exclusive events can never happen at the same time — rolling a 2 and rolling a 5 on one die roll, for instance. When that's true, there's no overlap to subtract, so P(A and B) is 0 and P(A or B) simplifies to plain addition: P(A) + P(B).
How do I find P(A and B) if I don't already know it?
It depends on the relationship between A and B. If they're independent, this site's and-probability calculator multiplies P(A) x P(B) to get it. If B's occurrence changes A's odds (or vice versa), use the conditional-probability calculator to work out the dependent joint probability instead.
Can the result go above 100% or below 0%?
It can, if the numbers entered aren't internally consistent with a real probability scenario — for example, if P(A and B) is set larger than either individual probability, which is mathematically impossible for genuine probabilities. A result outside 0-100% is a signal to double-check your inputs, particularly that P(A and B) doesn't exceed the smaller of P(A) and P(B).
What's the difference between this and the and-probability calculator?
This one answers 'at least one of A or B' by adding and correcting for overlap. The and-probability calculator answers a different question — 'both A and B together' — by multiplying, and only for independent events. They're complementary tools that often get used one after the other: find P(A and B) first, then feed it in here.
Does it matter which event I call A and which I call B?
No. Addition is symmetric, so P(A) + P(B) - P(A and B) gives the identical result whichever way round you label the two events — swapping A and B changes nothing about the outcome.