Combine two independent events, find the chance something happens at least once in n tries, or get exact binomial probabilities like P(X = k) and P(X ≥ k).
The rules for two independent events
Two events are independent when one happening doesn't change the chance of the other, like two coin flips.
P(A and B) = P(A) × P(B)
P(A or B) = P(A) + P(B) − P(A) × P(B)
P(exactly one) = P(A) + P(B) − 2 × P(A) × P(B)
P(neither) = (1 − P(A)) × (1 − P(B))
With P(A) = 0.5 and P(B) = 0.4: both happen with probability 0.2, at least one with 0.7, exactly one with 0.5, and neither with 0.3.
At least once in n tries
It's easier to find the chance it never happens, then subtract from 1.
P(at least once) = 1 − (1 − p)ⁿ
A 1% chance per try, over 100 tries: 1 − 0.99¹⁰⁰ = 0.634, or 63.4%. That's well short of 100%, which is what simply adding 1% a hundred times would suggest. The calculator also shows how many tries you need to reach 50%, 90%, 95% and 99%.
Binomial probability
For n independent trials, each with the same chance of success p, the chance of exactly k successes is:
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ
Ten fair coin flips with exactly 5 heads: C(10, 5) = 252, and 252 × 0.5¹⁰ = 0.2461. The calculator adds up every term for P(X ≤ k) and P(X ≥ k). It works with logarithms of the factorials, so it stays exact to about 12 digits even for 100,000 trials, where the numbers involved would overflow a normal calculator. For counting C(n, k) exactly, use the permutation and combination calculator.
When these formulas don't apply
- Dependent events. Drawing cards without putting them back changes the odds each time. Then P(A and B) = P(A) × P(B given A), not P(A) × P(B).
- Mutually exclusive events (they can't both happen) aren't independent. For them P(A or B) is just P(A) + P(B).
- The binomial needs a fixed number of trials, two outcomes each, the same p every time and independent trials.
Sources
How we checked this
Before every update, this calculator is run in a real browser with these known answers, and it must match all of them:
- Fraction 1/6 accepted: 4 tries = 1 − (5/6)⁴ = 51.7747% ✓
- P(A) = 0.5, P(B) = 0.4: P(A or B) = 0.7 ✓
- 30% and 0.6 mixed: P(A and B) = 0.18 ✓
- 1% chance, 100 tries: 63.396766% at least once ✓
- 10 fair coin flips, exactly 5 heads = 0.24609375 ✓
- 1,000 flips, exactly 500 heads = 0.02522501818 ✓
- p = 0: P(X = 0) = 1 ✓
Last checked October 26, 2026. Spotted something wrong? The math is done in your browser with the formula shown on this page, so you can check any result by hand.
Questions people ask
How do you calculate the probability of A and B?
For independent events, multiply: P(A and B) = P(A) × P(B). For dependent events, multiply P(A) by the chance of B once A has happened.
How do you calculate the probability of A or B?
Add the two and subtract the overlap: P(A) + P(B) − P(A and B). For independent events the overlap is P(A) × P(B).
What are the odds of something happening at least once?
Work out 1 − (1 − p)ⁿ. With a 1 in 6 chance per roll, the chance of at least one six in 4 rolls is 1 − (5/6)⁴ ≈ 51.8%.
What's the difference between P(X < k) and P(X ≤ k)?
P(X ≤ k) includes exactly k; P(X < k) doesn't. For counts, P(X < 5) is the same as P(X ≤ 4).
Can I enter probabilities as percents?
Yes. Type 0.3 or 30%. Any number above 1 is read as a percent, so 30 also means 30%.