Probability Calculator
Compute union, intersection, complement and conditional probabilities of events.
Inputs
Independent: knowing B occurred does not change P(A). Mutually exclusive: A and B cannot both occur.
Only used when relationship is set to 'Neither'.
Probability
0.760000
P(A ∩ B) used
0.240000
The joint probability of A and B both occurring.
Step by step
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∩ B)
P(A) × P(B) = 0.6 × 0.4
= 0.240000
Substitute
0.6 + 0.4 − 0.240000
= 0.760000
How it works
Probability quantifies how likely an event is on a scale from 0 (impossible) to 1 (certain). The addition rule gives P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive they cannot both occur, so P(A ∩ B) = 0. If they are independent, knowing B gives no information about A, so P(A ∩ B) = P(A) × P(B). Conditional probability P(A | B) = P(A ∩ B)/P(B) updates the likelihood of A given that B is known to have occurred.
Formulas
Addition rule
P(A or B) = P(A) + P(B) − P(A and B)
Independent events
P(A and B) = P(A) × P(B) when A and B are independent
Conditional probability
P(A given B) = P(A and B) / P(B)
Complement
P(not A) = 1 − P(A)
Frequently Asked Questions
What is the difference between mutually exclusive and independent events?
Mutually exclusive events cannot happen simultaneously — if A occurs, B cannot (P(A ∩ B) = 0). Independent events can both occur, but knowing one happened gives no information about the other (P(A | B) = P(A)). These are opposite extremes: two events with P(A) > 0 and P(B) > 0 cannot be both mutually exclusive and independent.
When is the addition rule P(A ∪ B) = P(A) + P(B)?
Only when A and B are mutually exclusive, so P(A ∩ B) = 0. For any other pair of events you must subtract P(A ∩ B) to avoid double-counting outcomes where both events occur.
Can a conditional probability be greater than the original probability?
Yes. If A and B are positively correlated, learning that B occurred increases the probability of A. For example, the probability of a positive cancer test is higher given that the patient actually has cancer than in the general population.
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