Calculate P(A or B), P(A and B) and P(A | B) for two events, or apply Bayes' theorem to a test result, with a probability tree and Venn diagram.
Cập nhật Ví dụ đã kiểm tra: 6
After seeing B, the probability of A moves from 1% to 9.17%. In natural frequencies: of 10,000 cases, about 90 have A and show B, while 891 show B without A — so only 90 of the 981 who show B actually have A.
For two events A and B, the addition rule gives P(A or B) = P(A) + P(B) − P(A and B), the multiplication rule gives P(A and B) = P(A) × P(B | A), and conditional probability is P(A | B) = P(A and B) / P(B). Bayes' theorem mode runs the conditional the other way: from a prior P(A) and the two evidence rates P(B | A) and P(B | not A), it returns the updated probability P(A | B).
The Bayes default is a screening test for a condition with 1% prevalence, 90% sensitivity and a 9% false-positive rate. A positive result raises the probability from 1% to 9.17%, because in 10,000 people the 90 true positives are outnumbered by 891 false positives.
Enter probabilities from 0 to 1; fractions such as 1/6 are accepted. Two-events mode rejects combinations that cannot exist, such as a P(A and B) larger than P(A) or P(B).
Nguồn đối chiếu: Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractions
Nguồn đối chiếu: P(A)P(B) = 0.2; addition rule 0.5 + 0.4 − 0.2
Nguồn đối chiếu: Definition: P(A ∩ B) = 0; union is the plain sum
Nguồn đối chiếu: Hand calculation with Python fractions: 0.4/0.5, 0.4/0.6 = 2/3, 1 − 0.7
Add the two probabilities and subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). For independent events with P(A) = 0.5 and P(B) = 0.4, the overlap is 0.5 × 0.4 = 0.2, so P(A or B) = 0.7. For mutually exclusive events the overlap is 0, so the probabilities simply add: 0.3 + 0.45 = 0.75.
Independent events don't affect each other's chances, so P(A and B) = P(A) × P(B). Mutually exclusive events can't happen together, so P(A and B) = 0. Two events with non-zero probabilities can't be both: if A rules out B, knowing A happened changes the chance of B to 0. Two coin flips are independent; heads and tails on one flip are mutually exclusive.
Because false positives from the large healthy group can outnumber true positives from the small affected group. With 1% prevalence, 90% sensitivity and 9% false positives, only 9.17% of positives are true. At 10% prevalence the same test gives 0.09 / (0.09 + 0.081) = 52.6%. This probability is the test's positive predictive value, and it depends on prevalence as much as on accuracy.
Subtract the chance that it never happens from 1: P(at least one) = 1 − (1 − p)ⁿ for n independent tries with probability p each. The chance of at least one six in 4 rolls of a die is 1 − (5/6)⁴ = 671/1296, about 0.5177, just above even odds. With p = 0.01 you need 69 tries before the chance passes 50%.
Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 6. Ví dụ, “Screening test: 1% prevalence, 90% sensitivity, 9% false positives” được kiểm tra bằng Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractions.
Grinstead & Snell, Introduction to Probability (AMS), §4.1 Discrete conditional probability and Bayes' formula; Gigerenzer et al. (2007). Helping doctors and patients make sense of health statistics. Psychological Science in the Public Interest 8(2), 53–96.
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