The core idea
For P(B) > 0, P(A|B) = P(A∩B)/P(B): restrict attention to B and measure the share also in A. Independence means knowing one event does not change the probability of the other. Bayes’ rule reverses conditioning by accounting for the base rates.
1. Conditioning changes the reference set
Let A and B denote events, and A∩B mean that both occur. In P(A|B), the vertical bar is read “given B”. It tells you to use B as the reference group, so the denominator is P(B), not P(A). This definition requires P(B) > 0. For a finite set of equally likely outcomes, count the outcomes in both events and divide by the number in B. Conditioning is not necessarily a story about one event happening first. Knowing a learner’s course can change a probability about their earlier choices. Nor does conditioning necessarily describe a cause: restricting a dataset to people who attended a workshop does not simulate forcing everyone to attend. Keep the statistical comparison separate from a claim about an intervention.
Sources: OpenStax: Two Basic Rules of Probability ↗ · OpenStax: Data, Sampling, and Variation in Data and Sampling ↗
2. Worked case: the denominator changes the answer
Consider an invented record of 100 learners. Sixty attended a practice workshop (A); fifty passed a later assessment (B); forty-two did both. Complete the table: among attendees, 42 passed and 18 did not; among the 40 non-attendees, 8 passed and 32 did not. Then P(B|A) = 42/60 = 0.70, while P(A|B) = 42/50 = 0.84. Both use the same overlap, but answer different questions. The overall pass probability is 50/100 = 0.50. Workshop attendance and passing are associated in this fictional record, but differences in prior preparation, time or motivation could explain some of the association. The table alone does not prove the workshop caused the difference. Check every row and column total before interpreting the fractions.
Sources: OpenStax: Contingency Tables ↗ · OpenStax: Data, Sampling, and Variation in Data and Sampling ↗
3. Independence is stronger than a casual impression
A and B are independent when P(A∩B) = P(A)P(B). When the conditioning probability is positive, this is equivalent to P(A|B) = P(A). In the workshop record, P(A∩B) = 0.42, whereas P(A)P(B) = 0.60×0.50 = 0.30. They are not independent. Mutually exclusive events cannot occur together, so their intersection has probability zero. If both have positive probability, they cannot also be independent: their product would be positive. Seeing one then rules out the other. Independence must come from a justified random mechanism or a stated model, not from the fact that events have different names. Observed frequencies can suggest dependence, but finite samples fluctuate; exact equality in a small table is not a universal proof about a wider population.
Sources: OpenStax: Independent and Mutually Exclusive Events ↗
4. Multiply along a path; add separate paths
Rearranging the conditional definition gives P(A∩B) = P(A)P(B|A). This multiplication rule always uses the conditional second factor; replacing it with P(B) requires independence. Imagine a box with three red and two blue counters, each equally likely to be drawn. Without replacement, two reds have probability (3/5)(2/4) = 3/10. With replacement and a fresh random draw, the probability is (3/5)(3/5) = 9/25. In a tree, branches leaving one point must sum to one, and a complete path represents an intersection. When several disjoint paths lead to the desired event, add their probabilities. For example, “one red and one blue” has two ordered paths: red then blue, and blue then red. Forgetting one path counts an order-specific event instead of the intended event.
Sources: OpenStax: Two Basic Rules of Probability ↗ · OpenStax: Independent and Mutually Exclusive Events ↗
5. Worked case: a good detector can still produce many false alarms
Use an invented audit of 1,000 components: 20 are defective and 980 are sound. A detector flags 18 defective components and misses 2; it also flags 49 sound components and leaves 931 unflagged. Among defective items, the flag rate is 18/20 = 90%. Among sound items, the false-flag rate is 49/980 = 5%. Yet among all 67 flagged components, only 18 are defective: P(defective|flagged) = 18/67 ≈ 26.87%. There are many more sound components available to generate false alarms. The 90% figure answers the reverse conditional question and is not the probability a flagged item is defective. This audit illustrates denominators; it is not performance evidence for any real detector. Changing the proportion of defective items can change the result even if both detection rates remain constant.
Choose the flagged column as the reference group
| Actual condition | Flagged | Not flagged | Total |
|---|---|---|---|
| Defective | 18 | 2 | 20 |
| Sound | 49 | 931 | 980 |
| Total | 67 | 933 | 1000 |
Sources: OpenStax: Contingency Tables ↗ · OpenStax: Two Basic Rules of Probability ↗
6. Bayes’ rule makes the reversal explicit
Write D for defective and F for flagged. The flagged group contains two disjoint paths, so P(F) = P(F|D)P(D)+P(F|not D)P(not D). In the audit this is 0.90×0.02+0.05×0.98 = 0.067. Dividing the first path by this total gives Bayes’ rule: P(D|F) = P(F|D)P(D)/P(F) = 0.018/0.067. The base rate P(D) is essential. The complement rule also works inside a conditioned group: P(not D|F) = 1−P(D|F), about 73.13%. Every term must refer to the same target population and compatible definitions. Rates measured in a different setting may not transfer. A reasoned probability update uses the evidence together with its background frequencies; it does not simply replace an earlier probability with a detector’s advertised percentage.
Sources: OpenStax: Two Basic Rules of Probability ↗ · OpenStax: Contingency Tables ↗
PUT IT INTO PRACTICE
Practice: two draws without replacement
- A box contains four red and six blue counters. Draw two at random without replacement. Define R₁ and R₂ as red on the first and second draw.
- Calculate P(R₂|R₁) and P(R₁∩R₂). Then find the probability of blue first and red second.
- Add both routes to a red second draw. Compare P(R₂|R₁) with the unconditional P(R₂) to assess independence.
- Solution: P(R₂|R₁) = 3/9 = 1/3; both red = (4/10)(3/9) = 2/15. Blue then red = (6/10)(4/9) = 4/15. Therefore P(R₂) = 6/15 = 0.4. Since 1/3 differs from 0.4, the events are dependent. The second draw’s overall red probability still equals the original red fraction.
Check your understanding
What fixes the denominator in P(A|B)?
The event after the bar, B. It is the population of outcomes being considered after conditioning.
Can P(A|B) equal P(B|A)?
Yes in some cases, but no general symmetry rule makes them equal. Their denominators usually differ.
Why are mutually exclusive positive-probability events dependent?
Learning that one occurred makes the other impossible, changing its probability from positive to zero.
Why did only about 27% of flagged components prove defective?
Sound components were far more numerous, so even their small false-flag rate generated more flags than true defects did.
When may we multiply unconditional probabilities?
When the relevant events are independent under the model. Otherwise use a conditional factor for the later branch.
Does conditioning on attendance measure a causal effect?
No. Attendees and non-attendees may differ beforehand. A causal interpretation needs additional design or assumptions.