Probability calculator (events & Bayes' theorem)

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.

更新于 2026年9月28日已验证的示例：6

P(A | B)

最大小数位数：6；取最近值，等距时远离零

P(A and B): 0.009
P(A or B): 0.0991
P(B): 0.0981
P(A | not B): 0.001109
P(not A): 0.99
P(not B): 0.9019
P(neither A nor B): 0.9009
P(exactly one of A, B): 0.0901
Independent?: No

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.

Probability tree

计算方法 S

Total probability of the evidence
$P(B) = P(B \mid A)P(A) + P(B \mid \bar A)P(\bar A) = 0.9 \times 0.01 + 0.09 \times 0.99 = 0.0981$
Bayes' theorem
$P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)} = \frac{0.009}{0.0981} = 0.091743$
Addition rule
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.01 + 0.0981 - 0.009 = 0.0991$
Complements
$P(\bar A) = 0.99,\quad P(\bar B) = 0.9019,\quad P(\overline{A \cup B}) = 0.9009$
Independence check
$P(A)P(B) = 0.000981 \ne P(A \cap B) = 0.009$

关于Probability calculator (events & Bayes' theorem)

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).

计算示例

Screening test: 1% prevalence, 90% sensitivity, 9% false positives

Calculate: Bayes' theorem
Prior probability P(A): 0.01
P(B | A): 0.9
P(B | not A): 0.09

P(A | B): 0.091743
P(B): 0.0981

核验来源：Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractions

Independent events 0.5 and 0.4

Calculate: Two events
P(A): 0.5
P(B): 0.4
How A and B relate: Independent

P(A and B): 0.2
P(A or B): 0.7
P(A | B): 0.5
P(neither A nor B): 0.3
Independent?: Yes

核验来源：P(A)P(B) = 0.2; addition rule 0.5 + 0.4 − 0.2

Mutually exclusive 0.3 and 0.45

Calculate: Two events
P(A): 0.3
P(B): 0.45
How A and B relate: Mutually exclusive

P(A and B): 0
P(A or B): 0.75
P(A | B): 0
P(exactly one of A, B): 0.75
Independent?: No

核验来源：Definition: P(A ∩ B) = 0; union is the plain sum

Known joint probability

Calculate: Two events
P(A): 0.6
P(B): 0.5
How A and B relate: I know P(A and B)
P(A and B): 0.4

P(A or B): 0.7
P(A | B): 0.8
P(B | A): 0.666667
P(neither A nor B): 0.3
P(exactly one of A, B): 0.3
Independent?: No

核验来源：Hand calculation with Python fractions: 0.4/0.5, 0.4/0.6 = 2/3, 1 − 0.7

常见问题

How do you calculate the probability of A or B?

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.

What is the difference between independent and mutually exclusive events?

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.

Why does a positive result on an accurate test often mean you probably don't have the condition?

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.

How do you calculate the probability of at least one event happening?

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%.

“Probability calculator (events & Bayes' theorem)”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。已按独立来源核验的计算示例：6。例如，“Screening test: 1% prevalence, 90% sensitivity, 9% false positives”根据Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractions进行核验。

这种方法出自哪里？

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.

关于此计算器

P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B \mid A)P(A) + P(B \mid \bar A)P(\bar A)},\qquad P(A \cup B) = P(A) + P(B) - P(A \cap B)

来源

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

已对照来源验证

此计算器包含 6 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。

Probability tree

Total probability of the evidence

Bayes' theorem

Addition rule

Complements

Independence check