CalcOpenly
한국어

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.

업데이트 검증한 예제: 6

How likely A is before seeing the evidence, e.g. how common a condition is.
Chance of seeing the evidence when A is true, e.g. a test's sensitivity.
Chance of seeing the evidence when A is false, e.g. the false-positive rate (1 − specificity).
시도하기
P(A | B)
P(A | B): 0.091743
최대 소수 자릿수: 6; 가장 가까운 값, 중간값은 0에서 먼 쪽으로
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

0.010.99Anot A0.90.10.090.91A and B: 0.009A and not B: 0.001not A and B: 0.0891not A, not B: 0.9009
계산 방법 S

  1. Total probability of the evidence

    P(B)=P(B∣A)P(A)+P(B∣Aˉ)P(Aˉ)=0.9×0.01+0.09×0.99=0.0981P(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

  2. Bayes' theorem

    P(A∣B)=P(B∣A)P(A)P(B)=0.0090.0981=0.091743P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)} = \frac{0.009}{0.0981} = 0.091743

  3. Addition rule

    P(A∪B)=P(A)+P(B)−P(A∩B)=0.01+0.0981−0.009=0.0991P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.01 + 0.0981 - 0.009 = 0.0991

  4. Complements

    P(Aˉ)=0.99,P(Bˉ)=0.9019,P(A∪B‾)=0.9009P(\bar A) = 0.99,\quad P(\bar B) = 0.9019,\quad P(\overline{A \cup B}) = 0.9009

  5. Independence check

    P(A)P(B)=0.000981≠P(A∩B)=0.009P(A)P(B) = 0.000981 \ne P(A \cap B) = 0.009

Probability calculator (events & Bayes' theorem) 소개

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).

계산 예제

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

검증 출처: 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

검증 출처: 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

검증 출처: 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

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

자주 묻는 질문

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)”의 정확도는 어느 정도인가요?

정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 6. 예를 들어 “Screening test: 1% prevalence, 90% sensitivity, 9% false positives”은 Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with Python fractions와 대조해 확인합니다.

이 계산 방법의 출처는 무엇인가요?

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.

이 계산기 소개

P(A∣B)=P(B∣A) P(A)P(B∣A)P(A)+P(B∣Aˉ)P(Aˉ),P(A∪B)=P(A)+P(B)−P(A∩B)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)

출처

  1. Grinstead & Snell, Introduction to Probability (AMS), §4.1 Discrete conditional probability and Bayes' formula
  2. Gigerenzer et al. (2007). Helping doctors and patients make sense of health statistics. Psychological Science in the Public Interest 8(2), 53–96

출처와 대조하여 검증

이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 6개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.

관련 계산기