Two kinds of probability question
A probability is a number from 0 (impossible) to 1 (certain), often written as a percentage from 0% to 100%. Choose how you want to type it — a probability of 0.25 and 25% are the same thing.
Two events combines the chances of two separate things, A and B. Enter P(A) and P(B) to see the chance that both happen, that at least one happens, that exactly one happens, that only A or only B happens, or that neither does. The table under the result lays all four combinations out as a grid.
Repeated trials is for doing the same thing several times — flipping a coin, rolling a die, sending out applications — where each attempt succeeds with the same probability p. Enter p, the number of trials n and a number of successes k to see the chance of exactly k, at least k, at most k, and at least one success. The chart shows the whole distribution.
The independence assumption
Both modes assume independence: the outcome of one event or trial doesn’t change the chances of another. Coin flips, dice rolls and draws where the card is put back are independent. Many real-world events are not: rain today makes rain tomorrow more likely, and drawing a card without putting it back changes what is left in the deck. For dependent events these formulas give the wrong answer, and you need a conditional probability instead (see below).
Probability formulas
For two independent events A and B:
P(A and B) = P(A) × P(B)
P(A or B) = P(A) + P(B) − P(A) × P(B)
P(neither) = (1 − P(A)) × (1 − P(B)) = 1 − P(A or B)
For n independent trials with success probability p, the number of successes X follows the binomial distribution:
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ
P(X ≥ 1) = 1 − (1 − p)ⁿ
- C(n, k)
- the number of ways to choose which k of the n trials are successes: n! ÷ (k! × (n − k)!)
- pᵏ × (1 − p)ⁿ⁻ᵏ
- the chance of one particular order of k successes and n − k failures
- P(X ≥ k)
- the sum of P(X = i) for every i from k to n; P(X ≤ k) sums from 0 to k
The expected number of successes is n × p and the standard deviation is √(n × p × (1 − p)). Binomial coefficients and powers become astronomically large or small as n grows, so the calculator works with logarithms and adds the tail probabilities term by term. Results are accurate to about nine significant digits up to 100,000 trials, and probabilities far smaller than 10⁻³⁰⁰ are still reported.
Worked examples
Two job applications. You estimate a 60% chance of an interview at one company and 30% at an unrelated one, and treat them as independent.
- Both: 0.6 × 0.3 = 0.18, so 18%.
- At least one: 0.6 + 0.3 − 0.18 = 0.72, so 72%.
- Neither: (1 − 0.6) × (1 − 0.3) = 0.4 × 0.7 = 0.28, so 28% — and 72% + 28% = 100%.
- Exactly one: 0.6 + 0.3 − 2 × 0.18 = 0.54, so 54% (42% only the first, 12% only the second).
Ten coin flips. With p = 0.5, n = 10 and k = 5:
- Ways to get 5 heads: C(10, 5) = 10! ÷ (5! × 5!) = 252.
- Chance of any one order: 0.5⁵ × 0.5⁵ = 0.5¹⁰ = 1/1,024 ≈ 0.000977.
- Exactly 5 heads: 252 ÷ 1,024 = 24.61% — about 1 in 4.06.
- At least 5 heads: (252 + 210 + 120 + 45 + 10 + 1) ÷ 1,024 = 638 ÷ 1,024 = 62.30%.
- At least one head: 1 − 0.5¹⁰ = 1,023 ÷ 1,024 = 99.90%.
When events aren’t independent
If B depends on A, replace P(B) with the conditional probability P(B | A), the chance of B once A has happened: P(A and B) = P(A) × P(B | A). Drawing two aces from a 52-card deck without putting the first back: 4/52 × 3/51 = 1/221, about 0.45%. With the card put back the draws are independent and the chance is 4/52 × 4/52 = 1/169, about 0.59%.
- Independent is not the same as mutually exclusive. Mutually exclusive events can’t happen together (a single die showing both 2 and 5), so P(A and B) = 0 and P(A or B) = P(A) + P(B). Two events with non-zero chances can’t be both independent and mutually exclusive.
- Past results don’t change independent odds. After five heads in a row the next flip is still 50% heads. The run was unlikely before it started, not the next flip.
- Small chances add up over many tries. A 1% risk per attempt becomes a 63.4% chance of at least one occurrence over 100 independent attempts: 1 − 0.99¹⁰⁰.
For the spread of measured data rather than chances, use the standard deviation calculator; for the mean of a set of results, the average calculator. Probabilities given as fractions such as 3/8 can be converted with the fraction calculator.
Frequently asked questions
What does it mean for events to be independent?
Two events are independent when knowing that one happened doesn’t change the probability of the other. Separate coin flips are independent; drawing two cards from a deck without putting the first back is not, because the first draw changes what is left.
How do I find the probability of at least one success?
Work out the chance of no successes and subtract it from 1: P(at least one) = 1 − (1 − p)ⁿ. The chance of at least one six in four rolls of a die is 1 − (5/6)⁴ = 671/1,296, about 51.8%.
What is the difference between “A or B” and “exactly one”?
“A or B” includes the case where both happen; “exactly one” excludes it. With P(A) = 60% and P(B) = 30%, P(A or B) is 72% but exactly one is 54%, because the 18% chance of both is left out.
Can I use this for events that are not independent?
Not directly. For dependent events, multiply P(A) by the probability of B given that A has happened. Two aces in a row without replacement is 4/52 × 3/51 = 1/221, not 4/52 × 4/52.
Why is exactly 50 heads in 100 flips so unlikely?
Although 50 is the single most likely result, there are 101 possible counts and many of them near 50 are almost as likely. The chance of exactly 50 is C(100, 50) ÷ 2¹⁰⁰, about 7.96%. The chance of between 40 and 60 heads is over 96%.
Sources
Last reviewed September 15, 2026