SEO 타이틀 (52): 확률 계산기 - 두 사건, 베이즈 및 시리즈
확률 계산기는 두 가지 사건 A와 B를 함께 처리합니다: P(A)와 P(B)를 입력하고, 이들이 독립적인지 P(A)∩ B를 나타내는지를 명시하며, 보완 확률, 교차점, 합집합, 배타 또는 확률, 조건부 확률을 읽습니다. 별도의 모드는 베이즈 정리, 반복 시도 시리즈, 그리고 정규 곡선 하의 영역을 다룹니다.
여기서 확률은 0와 1 사이의 숫자(또는 0%와 100%) 사이의 숫자입니다. 계산기는 두 형태를 모두 보여주므로 0.2와 20%의 결과는 결코 일치하지 않습니다.
두 사건의 확률을 계산합니다
P의 A와 P의 B를 소수점 또는 백분율로 입력합니다. 독립 사건의 경우 계산기가 교차점을 곱으로 제공합니다. 의존 사건의 경우, 교차점이나 조건문을 명시적으로 입력하여 이후 공식이 데이터의 이야기와 일치하도록 합니다.
0에서 1 밖의 확률, 혹은 두 사건보다 큰 교차점의 경우, 숫자를 만들어내는 대신 메시지로 멈춥니다.
유효하지 않은 입력(0-1 밖의 확률 또는 두 사건보다 큰 교차점)은 숫자를 만들어내는 대신 메시지로 중단됩니다.
보집합, 교차점, 합집합을 찾는다
A의 여집합은 1에서 P를 뺀 값이다. 교집합은 두 집합이 모두 발생할 확률이다; 독립성 하에 각각 개별 확률들의 곱과 같다. 합집합은 A의 P에 B의 P를 더한 값에서 교차를 뺀 값으로, 중복이 두 번 계산되는 것을 막는다.
이 세 가지 양은 대부분의 2 사건 숙제 문제의 기본 구성 요소입니다. P(A') = 1 − P(A) P(A ∩ B) = P(A) × P(B) (독립적인 경우) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ```
The complement is "not A." The intersection is "both." The union is "either A or B or both." Subtracting the intersection in the union formula stops double-counting the overlap.
Calculate exclusive or and mutually exclusive events
Exclusive or is the probability that exactly one of the two events occurs: P of A plus P of B minus twice the intersection. Mutually exclusive events cannot both occur, so their intersection is zero and the union simplifies to a plain sum.
Mutual exclusivity is a strong claim about the data; do not set the intersection to zero unless that claim is true.
P(A XOR B) = P(A) + P(B) − 2P(A ∩ B)Mutually exclusive events cannot both occur, so P(A ∩ B) = 0 and the union simplifies to P(A) + P(B). Mutually exclusive is stronger than "dependent"; check that claim against the story of the data before setting intersection to zero.
Calculate conditional probability
P of A given B equals the intersection divided by P of B. Read it as the updated probability of A after learning that B occurred. If A and B are independent, the conditional equals the original P of A.
Division by zero is blocked when P of B is zero, because conditioning on an impossible event is undefined in this framework. P(A | B) = P(A ∩ B) / P(B) ```
P(A | B)를 B가 발생했을 때 A가 발생할 확률로 읽는다. 만약 A와 B가 독립적이라면, P(A | B) = P(A)이다. 이들이 의존적일 때, 조건화는 확률을 바꾸며, 이것이 이 공식의 핵심이다.
0으로 나누는 것은 P(B) = 0 때에 차단된다: 불가능한 사건에 대한 조건화가 정의되지 않았다.
베이즈 정리를 적용한다
베이즈 정리는 새로운 증거 이후의 사전 확률을 그 증거의 우도를 사용하여 갱신한다. B가 주어진 후 A의 사후 P는 A x A의 이전 P가 주어진 B의 우도 P와 같으며, 모두 P의 B로 나눈다.
아래 의료 검사 수치는 이전 질병 발생률이 낮음에도 불구하고 매우 정확한 검사가 여전히 약간의 후방 후반을 남길 수 있음을 보여줍니다.
P(A | B) = [P(B | A) × P(A)] / P(B) ```
Rewrite P of B with the law of total probability when only conditionals are known: P of plus equals P of plus given disease times P of disease plus P of plus given no disease times P of no disease. Skipping that expansion and dividing by the prior alone is a common Bayes arithmetic error.
Prior P(A), likelihood P(B | A), and the marginal P(B) combine into the posterior P(A | B).
Medical test sketch. Disease prior P(D) = 0.01. Test sensitivity P(+|D) = 0.99. False positive rate P(+|no D) = 0.05. P(+) = 0.99×0.01 + 0.05×0.99 = 0.0099 + 0.0495 = 0.0594. P(D|+) = (0.99×0.01) / 0.0594 ≈ 0.167.
A positive test raises the chance of disease from 1% to about 16.7%, not to 99%. That gap is why Bayes belongs on a probability page rather than as a footnote.
Calculate probability across a series of events
For n independent trials with success probability p, the chance all n succeed is p to the n, and the chance of at least one success is one minus one minus p to the n. The at-least-one form usually wins by hand because the complement of zero successes is a single power rather than a long binomial sum across k.
P(all n 성공) = pⁿ
P(최소 한 번의 성공) = 1 − (1 − p)ⁿAt least one is usually easier through the complement: compute the chance of zero successes, then subtract from 1. For p = 0.1 and n = 10, P(at least one) = 1 − 0.9¹⁰ ≈ 1 − 0.3487 = 0.6513.
Calculate a probability of 0.5 and 0.4 for two events
Assume independence with P of A equal to 0.5 and P of B equal to 0.4. Intersection is 0.20, union is 0.70, exclusive or is 0.50, and P of A given B returns 0.50, matching P of A as independence requires. Every figure in the table can be reproduced from the formulas in the sections above without the interactive tool.
| Quantity | Working | Result | |
|---|---|---|---|
| Intersection | 0.5 × 0.4 | 0.20 | |
| Union | 0.5 + 0.4 − 0.2 | 0.70 | |
| XOR | 0.5 + 0.4 − 2×0.2 | 0.50 | |
| Complement of A | 1 − 0.5 | 0.50 | |
| P(A\ | B) | 0.2 / 0.4 | 0.50 |
Independence makes P(A|B) equal P(A), which matches 0.50 here.
Read the Venn diagram
Two circles show A and B; the overlap is the intersection. Shaded regions track the quantity you selected (union shade covers both lobes; XOR shade covers the lobes without the overlap). When probabilities are supplied, circle areas are drawn proportionally where the geometry allows, so a tiny intersection looks tiny.
Use the diagram to catch impossible inputs visually. If the overlap looks larger than either circle's story allows, the numeric intersection is inconsistent. The diagram does not replace algebra for dependent events with messy conditionals; it supports the algebra.
Venn diagrams do not replace the formulas. They make double-counting visible, which is why the union subtracts the overlap once.
Calculate probability from a normal distribution
Enter mean, standard deviation and one or two bounds; the calculator standardises to z and returns the area under the normal curve. For mean 100 and standard deviation 15, the probability of a value below 115 uses z equals 1 and area about 0.8413.
Use this mode for continuous below, above or between questions; use the two-event mode for logical combinations.
Example: μ = 100, σ = 15, P(X < 115) → z = (115−100)/15 = 1 → area ≈ 0.8413. Use this mode for continuous "below / above / between" questions; use the two-event mode for logical combinations of discrete events.
Independence is an assumption
Independence must come from the design or from evidence, not from convenience. Drawing cards without replacement makes successive hearts dependent; coin flips on a fair coin are the usual independence example. Multiplying probabilities that are actually dependent understates or overstates joint risk depending on the direction of dependence.
When only conditional information is available, rebuild the intersection as P of B times P of A given B before computing a union. The calculator accepts that path so the same identity holds for both independent and dependent stories.
Frequently asked questions
What is the difference between intersection and union?
Intersection is both events; union is either event (including both). Union adds the probabilities and subtracts the intersection so the overlap is not counted twice.
When can I multiply P(A) and P(B)?
When A and B are independent. If they are dependent, you need P(A ∩ B) or a conditional probability from data or a model.
What does mutually exclusive mean?
The events cannot occur together, so P(A ∩ B) = 0. A card cannot be both a heart and a spade. Mutually exclusive events are dependent except in trivial edge cases.
How does Bayes' theorem help?
It updates a prior probability after seeing new evidence, using the likelihood of that evidence. Medical testing and spam filtering are standard applications.
How do I find the probability of at least one success in n trials?
Compute 1 − (1 − p)ⁿ for independent trials with success probability p. Directly summing the binomial terms is equivalent but slower by hand.
Can probabilities exceed 1?
No. If a formula produces a value outside 0-1, an input assumption is wrong (often an intersection larger than one of the events).
Is odds the same as probability?
No. Odds compare successes to failures (for example 1:4); probability is successes over total possibilities (0.2). The Odds Calculator converts between them.
Does the normal mode assume a bell curve?
Yes. Area translation uses the normal model. For clearly skewed data, treat the result as approximate or use a different distribution.
Summary
The Probability Calculator returns complements, intersections, unions, XOR results and conditionals for two events, with Bayes, series-trials and normal-area modes for related questions. Independent events with P(A) = 0.5 and P(B) = 0.4 give intersection 0.2, union 0.7 and XOR 0.5.
Multiply only under independence; subtract the overlap in every union; and use Bayes when a prior must be updated after new evidence. Pair with the Odds Calculator when the inputs arrive as ratios rather than probabilities.