Conditional probability restricts attention to the given event. Reference: OpenStax’s terminology overview.
A finite model with colour and shape
Hypothetical example: Imagine 12 distinct tokens, all equally likely to be selected. Their attributes are arranged as follows:
| Colour | Circle | Square | Total |
|---|---|---|---|
| Red | 4 | 2 | 6 |
| Blue | 1 | 5 | 6 |
| Total | 5 | 7 | 12 |
The ordinary probability of red is P(red) = 6/12 = 1/2. Nothing has been conditioned on, so all 12 tokens remain in the denominator.
Read the wording before choosing the denominator
For P(red given circle), the given group is circle. Remove all seven squares from consideration. Five circles remain, and four are red:
P(red | circle) = red circles ÷ all circles = 4/5.
For P(circle given red), the given group is red instead. Now the relevant group has six tokens, of which four are circles:
P(circle | red) = red circles ÷ all red = 4/6 = 2/3.
These answers differ because the conditioning is reversed. The shared count, four red circles, is the numerator in both cases, but the denominator follows the group after “given”.
A reliable counting procedure
For any similarly specified finite model, name the condition first. If the question is “A given B”, count every outcome in B for the denominator, then count the outcomes belonging to both A and B for the numerator. Thus the structure is P(A | B) = count(A and B) ÷ count(B), provided the outcomes are equally likely.
This calculation describes the stated hypothetical model only. It is not evidence about an actual audience, website, screen, draw, or future result. The key reading habit is simple: the denominator is not automatically the total population; it is the group selected by the condition.
For a different question about interpreting past observations, see our history-screen reading note. The hypothetical counts above are not observed results.