Probability Calculator
Enter any probability values and get P(A), P(A'), P(A∩B), P(A∪B), and P(AΔB) instantly — covering single events, complement (NOT), intersection (AND), union (OR), exclusive OR, and repeated independent trials with full step-by-step results.
$
n
Enter your values above to see the results.
Tips & Notes
- ✓P(A) + P(A') = 1 always, without exception. If P(a product passing quality control) = 0.92, then P(failing) = 0.08 — not 0.07, not 0.09. The two must sum to exactly 1.
- ✓Multiply probabilities for AND (independent events), add then subtract for OR. The most frequent arithmetic error is adding when you should multiply: P(heads twice) = 0.5 × 0.5 = 0.25, not 0.5 + 0.5 = 1.0.
- ✓Convert percentages to decimals before calculating. P(A) = 30% must be entered as 0.30. Running the formula with 30 instead of 0.30 will produce results that are 100× too large.
- ✓For "at least once in n trials", always use 1 − P(A')ⁿ rather than multiplying P(A) by n. P(at least one 6 in 6 rolls) = 1 − (5/6)⁶ = 0.6651, not 6 × (1/6) = 1.0 — which is impossible.
- ✓Probabilities can never be less than 0 or greater than 1. If your calculation produces 1.3 or −0.2, an error occurred — either in the inputs or in the operation applied. Re-check whether events are independent or mutually exclusive.
Common Mistakes
- ✗Using the independent AND formula on dependent events. Drawing 2 aces from a deck without replacement: P(first ace) = 4/52, P(second ace) = 3/51 — not 4/52 again. Using 4/52 × 4/52 = 0.00592 instead of the correct 4/52 × 3/51 = 0.00452 produces a 31% overestimate.
- ✗Adding probabilities for OR without subtracting the intersection. P(A ∪ B) = P(A) + P(B) only when events are mutually exclusive. For overlapping events, omitting − P(A ∩ B) double-counts outcomes that satisfy both conditions and inflates the result.
- ✗Confusing mutually exclusive with independent. Mutually exclusive means P(A ∩ B) = 0 — both cannot occur together. Independent means P(A ∩ B) = P(A) × P(B) — each occurs regardless of the other. Rolling a 2 and rolling a 5 on one die are mutually exclusive. Rolling a 2 on one die and a 5 on another are independent.
- ✗Multiplying probabilities across non-independent trials. If a basketball player has a 70% free-throw average, P(making 3 consecutive) = 0.7³ = 0.343 only if each shot is truly independent. Fatigue and pressure create dependencies the simple multiplication ignores.
- ✗Treating probability 0 as impossible and probability 1 as certain in all contexts. In continuous distributions, P(X = exactly 5.000) = 0 even though X = 5 can occur. Zero probability does not mean impossible — it means infinitely rare.
Probability Calculator Overview
Probability is the mathematical language of uncertainty — every time you assess a risk, weigh a chance, or ask "what are the odds?", you are doing probability. Whether the outcome is a coin flip, a medical test, a quality defect, or a weather forecast, the same core rules govern all of it. This calculator handles single-event probability, combined events (AND, OR, XOR), complements (NOT), and repeated independent trials — turning a process that trips up even careful thinkers into a reliable, step-by-step result.
Basic probability — favorable outcomes over total outcomes:
P(A) = Favorable outcomes ÷ Total possible outcomes
EX: Rolling a standard 6-sided die and getting a 4 → Favorable = 1, Total = 6 → P(4) = 1/6 ≈ 0.1667 (16.67%)Complement (NOT) — the probability that an event does NOT occur:
P(A') = 1 − P(A)
EX: P(rolling a 4) = 1/6 → P(NOT rolling a 4) = 1 − 1/6 = 5/6 ≈ 0.8333 (83.33%)Intersection (AND) — both events occur, for independent events:
P(A ∩ B) = P(A) × P(B)
EX: P(rolling a 4 on first die) = 1/6, P(rolling a 4 on second die) = 1/6 → P(both 4s) = 1/6 × 1/6 = 1/36 ≈ 0.0278 (2.78%)For dependent events (outcome of A affects B), use the conditional form:
P(A ∩ B) = P(A) × P(B|A)
EX: Bag of 10 marbles — 7 black, 3 blue. Draw blue first (no replacement): P(blue) = 3/10. Then P(black | blue drawn) = 7/9 → P(blue then black) = 3/10 × 7/9 = 21/90 ≈ 0.2333Union (OR) — at least one event occurs:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
EX: P(even number on die) = 3/6, P(multiple of 3) = 2/6, P(both: number 6) = 1/6 → P(even OR multiple of 3) = 3/6 + 2/6 − 1/6 = 4/6 ≈ 0.6667For mutually exclusive events (cannot both occur), P(A ∩ B) = 0, so the formula simplifies:
P(A ∪ B) = P(A) + P(B)
EX: P(rolling a 2) = 1/6, P(rolling a 5) = 1/6 → P(rolling a 2 OR a 5) = 1/6 + 1/6 = 2/6 ≈ 0.3333Exclusive OR (XOR) — one or the other, but not both:
P(A Δ B) = P(A) + P(B) − 2 × P(A ∩ B)
EX: P(A) = 0.65, P(B) = 0.349, P(A ∩ B) = 0.65 × 0.349 = 0.2269 → P(A XOR B) = 0.65 + 0.349 − 2 × 0.2269 = 0.999 − 0.4538 = 0.5452Repeated independent trials — probability of event occurring at least once in n trials:
P(at least one) = 1 − P(A')ⁿ
EX: P(rolling a 6 on one die) = 1/6 ≈ 0.1667. What is P(at least one 6 in 4 rolls)? → P(no 6 in 4 rolls) = (5/6)⁴ = 0.4823 → P(at least one 6) = 1 − 0.4823 = 0.5177 (51.77%)Summary of all operations — when to use each:
| Operation | Symbol | Formula | Use When |
|---|---|---|---|
| Basic probability | P(A) | Favorable ÷ Total | Counting outcomes from a defined sample space |
| Complement (NOT) | P(A') | 1 − P(A) | Finding the probability of failure, absence, or opposite |
| AND (Independent) | P(A ∩ B) | P(A) × P(B) | Both events must occur; outcomes don't affect each other |
| AND (Dependent) | P(A ∩ B) | P(A) × P(B|A) | Both events must occur; first outcome changes the second |
| OR (Inclusive) | P(A ∪ B) | P(A) + P(B) − P(A ∩ B) | At least one event occurs; events can overlap |
| OR (Mutually Exclusive) | P(A ∪ B) | P(A) + P(B) | At least one occurs; events cannot happen simultaneously |
| XOR (Exclusive OR) | P(A Δ B) | P(A) + P(B) − 2×P(A ∩ B) | Exactly one occurs, never both at the same time |
| At least once in n trials | P(≥1) | 1 − P(A')ⁿ | Repeated independent attempts at the same event |
Frequently Asked Questions
Mutually exclusive events cannot occur simultaneously — if one happens, the other is impossible. Rolling a 3 and a 5 on one die: P(3 AND 5) = 0. Independent events can both occur but neither affects the other's probability. Rolling a 3 on one die and a 5 on a separate die are independent: P(3) = 1/6 regardless of the second die. Key formula difference: mutually exclusive uses P(A) + P(B) for OR; independent uses P(A) × P(B) for AND.
Use the complement rule: P(A') = 1 − P(A). If a production line has a 3% defect rate, P(no defect) = 1 − 0.03 = 0.97. For multiple independent events all not occurring: P(none fail) = P(A') × P(B') × P(C'). Three machines each with 5% failure probability → P(none fail) = 0.95 × 0.95 × 0.95 = 0.857 (85.7%). The complement is especially useful for 'at least one' problems — compute 1 minus the probability none occur.
Multiply the individual probabilities: P(A ∩ B) = P(A) × P(B). Example: P(heads) = 0.5, P(rolling a 6) = 1/6 → P(heads AND rolling a 6) = 0.5 × 0.1667 = 0.0833 (8.33%). For three independent events: P(passing three tests each with 80% pass rate) = 0.8 × 0.8 × 0.8 = 0.512 (51.2%). Only applies when events are truly independent — the outcome of A has zero effect on B.
Use the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Example: in a 52-card deck, P(King) = 4/52, P(Heart) = 13/52, P(King of Hearts) = 1/52 → P(King OR Heart) = 4/52 + 13/52 − 1/52 = 16/52 ≈ 0.3077 (30.77%). If mutually exclusive, P(A ∩ B) = 0: P(rolling 1 OR 6) = 1/6 + 1/6 = 2/6 ≈ 0.3333. Never skip the subtraction step unless events cannot both occur.
A probability of 0.05 means the event occurs 5 times out of every 100 independent trials on average — not exactly once in 20 tries. In any specific 20 trials it might occur 0, 1, 2, or more times. Over a very large number of trials, the frequency converges to 5%. In statistics, 0.05 is the conventional significance threshold (p-value), meaning results at that level occur by random chance about 1 in 20 times — which is why a single significant result alone is not considered conclusive proof.
Use the complement: P(at least once in n trials) = 1 − P(A')ⁿ. Example: a security system has a 2% false alarm rate per day. P(at least one alarm in 30 days) = 1 − 0.98³⁰ = 1 − 0.5455 = 0.4545 (45.45%). Even with a 2% daily rate, there is nearly a 50% chance of at least one alarm over a month — a result that surprises most people. Never multiply P(A) × n as it overestimates and can exceed 1.