Probability Calculator
Two solvers in one. Enter the probabilities of two independent events to get every combination — both, either, exactly one, neither, complements, and conditionals — or switch modes to find the probability that a normal variable lands between two values.
Outcome breakdown
Independent events, four corners
When A and B are independent (one happening tells you nothing about the other), every question reduces to four corner outcomes: both happen with probability P(A)P(B), only A with P(A)(1 − P(B)), only B with (1 − P(A))P(B), and neither with (1 − P(A))(1 − P(B)) — the four always sum to 1. “At least one” is everything except the neither corner, and because of independence the conditionals collapse: P(A|B) is just P(A). In normal mode, the calculator converts each bound to a z-score and subtracts the two cumulative areas.
How it’s calculated
Independent events: P(A∩B) = P(A)P(B); P(A∪B) = P(A) + P(B) − P(A)P(B); P(exactly one) = P(A) + P(B) − 2P(A)P(B); P(neither) = (1 − P(A))(1 − P(B)); P(A|B) = P(A). Normal mode: P(a < X < b) = Φ((b − μ)/σ) − Φ((a − μ)/σ), with Φ the standard normal CDF computed via the error function.
Results update as you type and are for education, not professional advice — double-check any number that matters.
Worked example
With P(A) = 0.5 and P(B) = 0.4: both = 0.5 × 0.4 = 0.2; at least one = 0.5 + 0.4 − 0.2 = 0.7; exactly one = 0.5; neither = 0.5 × 0.6 = 0.3; only A = 0.3, only B = 0.2. Normal mode: for μ = 100, σ = 15, the probability of landing between 85 and 115 (z = −1 to +1) is 0.6827, with 0.1587 in each tail.
Common mistakes
- Adding P(A) + P(B) for “A or B” without subtracting the overlap — you double-count the both case.
- Using these formulas on dependent events — drawing cards without replacement is not independent.
- Entering percentages (40) where the tool expects decimals (0.4) — probabilities must be 0 to 1.
Where it is used
- Homework on unions, intersections, and complements.
- Reliability: chance that at least one of two independent backups works.
- Normal-model questions — test scores, heights, measurement error between two limits.
Frequently asked questions
What does independence actually mean?
Knowing one event happened does not change the probability of the other: P(A|B) = P(A). Two coin flips are independent; drawing two cards without replacement is not, because the first draw changes the deck.
How do I get P(A or B)?
Add the two probabilities, then subtract the both-happen overlap so it is not counted twice: P(A) + P(B) − P(A)P(B). For 0.5 and 0.4 that is 0.9 − 0.2 = 0.7.
What is the difference between at least one and exactly one?
At least one includes the case where both happen (0.7 in the example); exactly one excludes it (0.5). The outcome table in the steps panel shows all four corners summing to 1.
How does the normal mode compute its answer?
Each bound is standardized to z = (x − μ)/σ, then the area between the two z-values under the standard normal curve is Φ(z₂) − Φ(z₁), computed with the error function — the same math as z-tables, without the table.
Can I leave one normal bound open-ended?
Use a bound far outside the data instead — μ ± 6σ is effectively infinity (area beyond it is about 1 in a billion). For one-sided questions, the z-score calculator gives left and right tail areas directly.