Give the probabilities of two events and get every combination: both, either, neither, exactly one, and conditional probabilities.
| P(A and B) | 0.2000 |
|---|---|
| P(A or B) | 0.7000 |
| P(not A), P(not B) | 0.5000, 0.6000 |
| Exactly one of A, B | 0.5000 |
| Neither | 0.3000 |
| P(A | B) | 0.5000 |
| P(B | A) | 0.4000 |
Working: P(A or B) = 0.5 + 0.4 − 0.2000 = 0.7000
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A common slip is to add probabilities across repeated tries: twenty 5% chances must make 100%. They don’t, because once the event has happened, later tries can’t make it happen “again” in the at-least-once sense.
Set the chance per try and the number of tries. The dots show 100 people each making that many attempts; the filled ones succeeded at least once.
At least once: 64.2% · naive sum would say 100%
P(A∪B) = P(A) + P(B) − P(A∩B) · P(A|B) = P(A∩B) ÷ P(B)
Independent: P(A∩B) = P(A)·P(B). Mutually exclusive: P(A∩B) = 0. Complement: P(not A) = 1 − P(A). For the number of successes over many independent trials, move on to the binomial distribution calculator.
P(A and B) is the chance both happen. P(A or B) is the chance at least one happens, found with the addition rule P(A) + P(B) − P(A and B). Subtracting the overlap stops it being counted twice.
Events are independent if one happening does not change the chance of the other, so P(A and B) = P(A) × P(B). Two coin flips are independent. Drawing two cards without replacement is not.
Events that cannot happen together, such as rolling a 2 and a 5 on one die. Then P(A and B) = 0 and P(A or B) = P(A) + P(B). Mutually exclusive events with non-zero probability are never independent.
P(A | B) = P(A and B) ÷ P(B): the chance of A once you know B happened. If 10% of students take both French and Spanish and 40% take Spanish, P(French | Spanish) = 0.10 ÷ 0.40 = 0.25.
One minus the probability of no successes: 1 − (1 − p)ⁿ for independent tries. With a 5% chance each time, 20 tries give 1 − 0.95²⁰ ≈ 64%, not 100%.