- P(A) — probability of event A
- 0.5
- P(B) — probability of event B
- 0.4
- P(A∩B) — probability of both
- 0.2
0.7000
Open with these values0.7000
Result: 0.7000Adding two probabilities counts the overlap twice, so the overlap has to come off once. A 50 % event and a 40 % event that coincide 20 % of the time give 70 %, not 90 %. Enter 0 for the overlap when the two events cannot both happen.
0.7000
Open with these values0.8000
Open with these values0.7500
Open with these valuesP(A∪B) = P(A) + P(B) − P(A∩B)
Outcomes in which both events happen belong to A and to B alike. Adding P(A) and P(B) counts that shared region twice, and subtracting P(A∩B) removes the double count exactly once.
The subtraction falls away only when the two cannot both happen, leaving P(A∪B) = P(A) + P(B). For 0.3 and 0.5 that is 0.8000.
Two independent halves still overlap: P(A∩B) = 0.25, and their union is 0.7500, not 1. Entering 0 for that overlap would claim the two can never coincide.
The overlap cannot exceed the smaller of P(A) and P(B), and the union cannot exceed 1. Numbers that break those rules describe events that cannot exist, and the nearest valid probability is shown instead.
P(A) = 0.5 and P(B) = 0.4 make the chance of A or B 0.9.
That holds only if the two never coincide. With an overlap of 0.2 the union is 0.5 + 0.4 − 0.2 = 0.7.
The overlap only has to be subtracted for dependent events.
It has to come off whenever both events can happen together. Two independent halves overlap 0.25, and their union is 0.7500 rather than 1.
A union above 1 just means the events are very likely.
No probability exceeds 1, so those three numbers describe events that cannot exist. P(A) = 0.7 with P(B) = 0.6 needs an overlap of at least 0.3, which lands exactly on 1.
| P(A), P(B), P(A∩B) | Case | P(A∪B) |
|---|---|---|
| 0.5, 0.4, 0.2 | overlapping events | 0.7000 |
| 0.3, 0.5, 0 | mutually exclusive | 0.8000 |
| 0.25, 0.15, 0.05 | small overlap | 0.3500 |
| 0.5, 0.5, 0.25 | two independent halves | 0.7500 |
| 0.7, 0.6, 0.3 | heavy overlap, certain union | 1.0000 |
Use the addition rule P(A∪B) = P(A) + P(B) − P(A∩B): add the two probabilities and subtract the chance that both happen. With P(A) = 0.5, P(B) = 0.4 and P(A∩B) = 0.2 the union is 0.7.
Because the outcomes in which both events happen belong to A and to B alike. Adding P(A) and P(B) counts that shared region twice, and subtracting P(A∩B) removes the double count exactly once.
Then they cannot both occur, their intersection is 0, and the rule simplifies to P(A∪B) = P(A) + P(B). Rolling a 1 or a 2 on one die is 1/6 + 1/6 = 1/3.
It is the probability of the union of A and B: the chance that A happens, that B happens, or that both do. In plain language it is P(A or B), with an inclusive or that allows both at once.
The overlap cannot exceed the smaller of P(A) and P(B), and the union cannot exceed 1. Inputs that break those rules describe events that cannot exist, and the calculator falls back to the nearest valid probability rather than showing one above 1.
Information, not professional advice.
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