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.

更新日 2026年9月28日検証済みの例：6

P(A | B)

小数点以下の最大桁数：6；最も近い値へ、等距離ならゼロから遠い値へ

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

計算方法 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
$P(A)P(B) = 0.000981 \ne P(A \cap B) = 0.009$

Probability calculator (events & Bayes' theorem)について

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).

計算例

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

照合元：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

照合元：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

照合元：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

照合元：Hand calculation with Python fractions: 0.4/0.5, 0.4/0.6 = 2/3, 1 − 0.7

よくある質問

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)」の精度はどのくらいですか？

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。独立した出典の解答と照合した計算例：6。例えば、「Screening test: 1% prevalence, 90% sensitivity, 9% false positives」はGigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractionsと照合しています。

この計算方法の出典は何ですか？

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.

この計算機について

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)

出典

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

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 6 件あります。テストに組み込まれており、ここでも実行できます。

Probability tree

Total probability of the evidence

Bayes' theorem

Addition rule

Complements

Independence check