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.
By AdminDrafted with AI writing toolsUpdated Checked examples: 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. Of the 981 cases showing B, about 90 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).
When all outcomes are equally likely, P(event) = favorable outcomes ÷ total outcomes. An even number on a fair die has 3 favorable faces out of 6, so P = 3/6 = 0.5. Every probability lies between 0, impossible, and 1, certain; 0.25 means a 25% chance.
Two events combine through the addition, multiplication and conditional rules below. Bayes' theorem, P(A | B) = P(B | A) × P(A) ÷ P(B), turns a conditional probability around, for example from a test's accuracy to the meaning of a positive result.
Roll two dice, with A a six on the first and B a six on the second. Both happen with probability 1/6 × 1/6 = 1/36, and at least one six with 1/6 + 1/6 − 1/36 = 11/36, about 0.306.
With independent probabilities 0.5 and 0.4, both events happen with probability 0.2 and at least one with 0.7. Neither happens with 0.3, and exactly one with 0.5.
| Relationship | P(A) | P(B) | P(A and B) | P(A or B) | P(A given B) |
|---|---|---|---|---|---|
| Independent | 0.5 | 0.4 | 0.2 | 0.7 | 0.5 |
| Mutually exclusive | 0.3 | 0.45 | 0 | 0.75 | 0 |
| Known intersection | 0.6 | 0.5 | 0.4 | 0.7 | 0.8 |
| Known conditional | 0.6 | 0.5 | 0.3 | 0.8 | 0.6 |
In the last row the known input is P(B | A) = 0.5, so the multiplication rule gives P(A and B) = 0.6 × 0.5 = 0.3.
Independent events don't change each other's chances, so P(A and B) = P(A) × P(B): two coin flips, or two dice. Mutually exclusive events can't happen together, so P(A and B) = 0: heads and tails on the same flip.
Two events with positive probabilities can't be both. If A rules out B, learning that A happened drops B's chance to 0, the opposite of independence.
P(A | B) and P(B | A) are usually different. In the known-intersection row, P(A | B) = 0.4 ÷ 0.5 = 0.8, but P(B | A) = 0.4 ÷ 0.6 = 0.667.
An overlap also has limits. It can't exceed either probability, and it must be at least P(A) + P(B) − 1. With 0.6 and 0.5 it lies between 0.1 and 0.5, so a stated P(A and B) of 0.7 is impossible.
Bayes' theorem updates a prior probability with evidence. Its denominator is the total chance of the evidence: P(B) = P(B | A) P(A) + P(B | not A) P(not A).
Take a condition affecting 1% of people, a test that detects 90% of cases and a 9% false-positive rate, the screening example used by Gigerenzer and colleagues (2007). Out of 10,000 people:
The 90% detection rate is not the chance that a positive result is right, because the large group without the condition supplies most of the positives.
| Prior P(A) | P(B given A) | P(B given not A) | P(B) | Updated P(A given B) |
|---|---|---|---|---|
| 0.01 | 0.9 | 0.09 | 0.0981 | 0.0917431193 |
| 0.1 | 0.9 | 0.09 | 0.171 | 0.5263157895 |
| 0.5 | 0.8 | 0.2 | 0.5 | 0.8 |
| 0 | 0.9 | 0.09 | 0.09 | 0 |
At a 10% prior, the same test's positive result means a 52.6% chance, so the answer depends on prevalence as much as on accuracy. A prior of 0 stays 0 whatever the evidence. If the evidence can never occur, P(B) = 0 and the update is undefined.
For dice totals, coin runs and card draws, the dice, coin and card calculator works out the events directly. To count the outcomes first, use the permutation and combination calculator.
OpenStax explains probability terminology and the addition and multiplication rules. Chapter 4 of Grinstead and Snell's Introduction to Probability develops conditional probability and Bayes' formula.
Checked against: Gigerenzer et al. (2007) mammography example (about 1 in 10 positives has the disease); exact 0.009/0.0981 = 10/109 with exact fractions
Checked against: P(A)P(B) = 0.2; addition rule 0.5 + 0.4 − 0.2
Checked against: Definition: P(A ∩ B) = 0; union is the plain sum
Checked against: Hand calculation with exact 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%.
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