Probability Calculator

Calculate basic probability, conditional probability, joint probability, and union probability with step-by-step solutions.

For learning and homework help — verify critical calculations independently.

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Calculator

Select a calculation type and enter the required values to calculate probabilities with step-by-step solutions.

The short answer

Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain): P(A) = favorable outcomes ÷ total outcomes. For combining two events, use the right rule for how they relate: multiply for "and" with independent events, add and subtract overlap for "or," and divide for "given that" (conditional probability).

Key takeaways

  • P(A and B) = P(A) × P(B) only works when A and B are independent — dependent events need conditional probability instead.
  • P(A or B) = P(A) + P(B) − P(A and B) — the subtraction avoids double-counting outcomes that satisfy both events.
  • Conditional probability, P(A|B) = P(A and B) ÷ P(B), narrows the "total outcomes" down to just the cases where B already happened.
  • The complement rule — P(at least one) = 1 − P(none) — is often the fastest way to solve "at least one" problems that would otherwise need many separate cases added together.

The core probability rules

Rule Formula Used for
BasicP(A) = favorable/totalA single event
UnionP(A or B) = P(A)+P(B)−P(A∩B)Either event happens
Joint (independent)P(A and B) = P(A) × P(B)Both happen, unrelated events
ConditionalP(A|B) = P(A∩B)/P(B)One event given another already happened

Worked example: rolling at least one six

What's the probability of rolling at least one six in 4 rolls of a fair die?

P(no six on one roll) = 5/6

P(no six in 4 rolls) = (5/6)⁴ ≈ 0.482

P(at least one six) = 1 − 0.482 ≈ 0.518 (51.8%)

Trying to solve this by directly adding up "exactly one six," "exactly two sixes," and so on would take far more work — the complement rule turns a multi-case problem into a single subtraction.

Independent vs. dependent events

Two coin flips are independent — the first flip's result has zero effect on the second. But drawing two cards from a deck without replacement is dependent — if the first card is a heart, there's one fewer heart left for the second draw, which changes its probability. Whenever removing or using up part of the sample space changes the odds for what comes next, the events are dependent, and conditional probability (not simple multiplication) is the right tool.

Common mistakes to avoid

  • Adding P(A) + P(B) for "or" without subtracting the overlap P(A and B) — this double-counts outcomes that satisfy both events.
  • Multiplying P(A) × P(B) for events that aren't actually independent — that shortcut only applies when one event has no effect on the other.
  • Confusing P(A|B) with P(B|A) — these are generally different values, a mix-up behind many real-world misreadings of medical test results and statistics.
  • Ending up with a probability outside the 0-to-1 range — that always signals an error upstream, since a valid probability can never be negative or greater than 1.

Frequently Asked Questions

What is probability?

Probability measures how likely an event is, from 0 (impossible) to 1 (certain). A fair coin flip has probability 0.5 for heads.

What is the difference between theoretical and experimental probability?

Theoretical probability comes from a model — a die shows 1/6 per face. Experimental probability counts outcomes in trials: heads 47 times in 100 flips = 0.47.

How do you combine probabilities?

For independent events A and B, P(A and B) = P(A) × P(B). For mutually exclusive events, P(A or B) = P(A) + P(B).

What is conditional probability?

P(A given B) is the probability of A when B is known to have occurred: P(A|B) = P(A and B) / P(B).

How do I use this probability calculator?

Choose a scenario — single event, multiple trials, or combinations — enter parameters, and click Calculate to see probability and supporting steps.

How do you calculate the probability of rolling at least one six in several dice rolls?

Use the complement rule: P(at least one six) = 1 − P(no sixes at all). Rolling a fair six-sided die 4 times, P(no six on one roll) = 5/6, so P(no six in 4 rolls) = (5/6)⁴ ≈ 0.482. That makes P(at least one six) ≈ 1 − 0.482 = 0.518, or about 51.8% — a classic problem first studied by Pascal and Fermat in the 1650s.

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