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Formula
P(A and B) = P(A) × P(B); P(A or B) = P(A) + P(B) − P(A) × P(B); P(not A) = 1 − P(A); P(at least once) = 1 − (1 − p)ⁿ
- P(A), P(B)
- the probabilities of events A and B, from 0 (impossible) to 1 (certain)
- p
- the probability of the event on a single trial
- n
- the number of independent trials
What probability measures
A probability is a number from 0 to 1 describing how likely an event is. A fair coin landing heads has a probability of 0.5; rolling a six on a die has a probability of 1 ÷ 6 ≈ 0.1667. Multiplying by 100 turns a probability into a percentage, so 0.1667 is about a 16.67% chance.
This probability calculator covers the three questions people ask most often: the chance of a single event from a count of outcomes, how two independent events combine, and the chance that something happens at least once if you keep trying.
How to use the probability calculator
Pick the tab that matches your question.
- Two events: enter P(A) and P(B) as decimals. You get P(A and B), P(A or B), the chance of exactly one, the chance of neither, and P(not A) and P(not B).
- Single event: enter the number of favorable outcomes and the total number of equally likely outcomes. You get the probability, its complement, a “1 in N” frequency and the odds.
- At least once in n trials: enter the chance per trial and the number of trials. You get the chance of at least one occurrence, of none, of it happening every time, and the expected count.
How to calculate probability by hand
Two independent events
Flip a coin and roll a die. Let A be heads (0.5) and B be rolling a six (1/6 ≈ 0.166667).
- And: 0.5 × 0.166667 = 0.083333. There’s about an 8.33% chance of heads and a six together.
- Or: 0.5 + 0.166667 − 0.083333 = 0.583333. Subtracting the overlap stops the heads-and-six outcome being counted twice.
- Not: P(not A) = 1 − 0.5 = 0.5.
- Neither: (1 − 0.5) × (1 − 0.166667) = 0.416667. Note that P(neither) = 1 − P(A or B).
At least once
What’s the chance of rolling at least one six in four rolls? The chance of no six on one roll is 5/6, so the chance of no six in four rolls is (5/6)⁴ ≈ 0.482253. Subtract from 1: about 0.517747, or 51.77%. That’s slightly better than even, which is the basis of a famous 17th-century gambling bet.
How fast “at least once” adds up
The table shows the chance of at least one occurrence for an event with a 1% chance per trial.
| Trials (n) | P(at least once) |
|---|---|
| 1 | 1% |
| 10 | 9.56% |
| 25 | 22.22% |
| 50 | 39.5% |
| 100 | 63.4% |
| 200 | 86.6% |
| 300 | 95.1% |
| 500 | 99.34% |
Even after 100 tries at 1%, there’s still about a 37% chance of no success at all. Over many trials, the expected count n × p is a better guide to how many times the event will happen, while the at-least-once figure tells you how likely you are to see it even once.
Real-life uses
- Weather and travel. If there’s a 30% chance of rain on each of two days, the chance of at least one wet day is 1 − 0.7 × 0.7 = 51%, assuming the days are independent.
- Reliability. If each of three backup systems fails 2% of the time independently, all three fail together with probability 0.02³ = 0.000008.
- Games. Dice, card and lottery odds start from favorable ÷ total outcomes. For counts of hands or tickets, the combination calculator gives the total number of outcomes.
- Testing and screening. Running a test with a 5% false-alarm rate 10 times gives a 40% chance of at least one false alarm, which is why multiple comparisons need correction.
Common mistakes
- Adding when you should multiply. “And” for independent events multiplies; adding two probabilities only works for mutually exclusive events in an “or” question.
- Forgetting the overlap. P(A or B) = P(A) + P(B) is only true when A and B can’t both happen.
- Treating dependent events as independent. Drawing cards without replacement, or a patient’s two related symptoms, needs conditional probability.
- The gambler’s fallacy. After five heads in a row, the next flip of a fair coin is still 50%. Independent trials have no memory.
- Using percentages in the formula. Convert 40% to 0.4 before multiplying, or the result will be off by a factor of 100.
Glossary
- Complement: the event not happening; P(not A) = 1 − P(A).
- Independent events: events where one outcome doesn’t affect the other.
- Mutually exclusive: events that can’t happen together, so P(A and B) = 0.
- Odds: favorable outcomes compared with unfavorable ones, such as 1 : 5 for a six on a die.
For probabilities under a bell curve, such as the share of test scores above a cutoff, use the z-score calculator. To simulate draws, the random number generator picks numbers in any range.
Worked examples
Two independent events
A 60% chance of rain on Saturday and an independent 30% chance your flight is delayed.
Inputs (Two events): Probability of A, P(A) = 0.6; Probability of B, P(B) = 0.3
P(A and B) = P(A) × P(B) = 0.6 × 0.3 = 0.18 P(A or B) = P(A) + P(B) − P(A and B) = 0.6 + 0.3 − 0.18 = 0.72 P(not A) = 1 − 0.6 = 0.4 P(neither) = (1 − 0.6) × (1 − 0.3) = 0.28
Answer: For independent events with P(A) = 0.6 and P(B) = 0.3: P(A and B) = 0.18, P(A or B) = 0.72, P(neither) = 0.28.
Drawing an ace from a deck
One card drawn from a shuffled 52-card deck containing 4 aces.
Inputs (Single event): Favorable outcomes = 4; Total possible outcomes = 52
P(A) = favorable ÷ total = 4 ÷ 52 = 0.076923 P(not A) = 1 − 0.076923 = 0.923077
Answer: P(A) = 4 ÷ 52 = 0.076923 (7.6923%).
At least one success in 20 tries
An event with a 5% chance on each independent attempt, tried 20 times.
Inputs (At least once in n trials): Probability per trial, p = 0.05; Number of trials, n = 20
P(never) = (1 − p)ⁿ = (1 − 0.05)²⁰ = 0.358486 P(at least once) = 1 − 0.358486 = 0.641514
Answer: With p = 0.05 per trial, the chance of at least one occurrence in 20 trials is 0.641514 (64.1514%).
One event that cannot happen
Edge case: if P(B) = 0, A and B can never both happen, and P(A or B) equals P(A).
Inputs (Two events): Probability of A, P(A) = 0.45; Probability of B, P(B) = 0
P(A and B) = P(A) × P(B) = 0.45 × 0 = 0 P(A or B) = P(A) + P(B) − P(A and B) = 0.45 + 0 − 0 = 0.45 P(not A) = 1 − 0.45 = 0.55 P(neither) = (1 − 0.45) × (1 − 0) = 0.55
Answer: For independent events with P(A) = 0.45 and P(B) = 0: P(A and B) = 0, P(A or B) = 0.45, P(neither) = 0.55.
Assumptions and limitations
- The two-event and at-least-once modes assume the events are independent, meaning one happening doesn't change the chance of the other. For dependent events, multiply by the conditional probability instead.
- Probabilities are entered as decimals from 0 to 1. Enter 0.25 for 25%.
- The single-event mode assumes every outcome is equally likely, as with a fair die or a well-shuffled deck.
- Results are shown to 6 decimal places and as percentages to 4. Very small probabilities are shown in scientific notation.
- The at-least-once result is computed as −expm1(n × log1p(−p)), which stays accurate for tiny probabilities over millions of trials.
- Odds are shown in lowest terms as favorable to unfavorable outcomes, written a : b.
Frequently asked questions
What is the difference between "and" and "or" probability?
"And" is the chance both events happen, found by multiplying for independent events. "Or" is the chance at least one happens, found by adding the two probabilities and subtracting the overlap so it isn't counted twice. "Or" is always at least as large as either event alone, and "and" is never larger.
How do I calculate the probability of something happening at least once?
Find the chance it never happens and subtract from 1. If an event has a 10% chance per try, the chance it never happens in 5 tries is 0.9⁵ ≈ 0.59, so the chance of at least once is about 0.41, or 41%.
Why can't I just multiply the chance by the number of tries?
Because that double-counts the tries where the event happens more than once, and it can exceed 100%. Ten tries at 20% would give 200%, but the real chance of at least one success is 1 − 0.8¹⁰ ≈ 89%.
What makes two events independent?
Events are independent when the outcome of one gives no information about the other, such as two coin flips or two dice. Drawing two cards without putting the first back is not independent, because the first card changes what's left in the deck.
How do I convert a percentage or odds to a probability?
Divide a percentage by 100, so 35% becomes 0.35. Odds of a to b in favor become a ÷ (a + b), so odds of 1 to 4 equal a probability of 1 ÷ 5 = 0.2.
What does a probability of 0 or 1 mean?
A probability of 0 means the event can't happen, and 1 means it's certain. Everything else falls in between, so a result outside 0 to 1 always signals a mistake in the working.
Sources
- 3.2 Independent and Mutually Exclusive Events (Introductory Statistics 2e) — OpenStax, Rice University. Checked October 12, 2026.
- 3.3 Two Basic Rules of Probability (Introductory Statistics 2e) — OpenStax, Rice University. Checked October 12, 2026.
- 3.1 Terminology (Introductory Statistics 2e) — OpenStax, Rice University. Checked October 12, 2026.