Why can a reverse conditional probability be different from the original one?
Quick answer
Reversing a conditional probability changes which group you start with. For P(A|B), divide the number of cases in both A and B by the total number of B cases. For P(B|A), divide that same overlap by the total number of A cases [S1]. For example, if 10 of 20 B cases are also A, but 10 of 40 A cases are also B, the probabilities are 10/20 = 50% and 10/40 = 25%.
The question, the details, the next step.
Why aren’t P(A given B) and P(B given A) the same? I keep thinking I can just turn one around.
Because the condition tells you which group to look inside. In P(A|B), you start with the B cases; in P(B|A), you start with the A cases. The shared cases appear in both calculations, but the denominator changes: P(A|B) = P(A and B) / P(B), while P(B|A) = P(A and B) / P(A) [S1].
For a small, invented example, imagine 100 people. Let A mean “reads comics” and B mean “plays chess.” Suppose 10 do both, 30 read comics but don’t play chess, 10 play chess but don’t read comics, and 50 do neither. These constructed counts are just for illustrating the calculation, not real-world data.
Can you lay out the counts so I can see exactly which denominator to use?
Here are the counts. Each person is in exactly one category:
- Both reads comics (A) and plays chess (B): 10
- Reads comics, but does not play chess: 30
- Plays chess, but does not read comics: 10
- Neither: 50
The A total is 10 + 30 = 40. The B total is 10 + 10 = 20. The overlap—the people in both A and B—is 10. The total is 100. Keep these three counts visible: overlap = 10, B total = 20, and A total = 40.
Show both calculations side by side. I want the denominator to stay visible.
For “reads comics given that the person plays chess,” the condition is B. Use the total number of chess players as the denominator:
P(A|B) = overlap / B total = 10 / 20 = 50%.
For “plays chess given that the person reads comics,” the condition is A. Use the total number of comic readers as the denominator:
P(B|A) = overlap / A total = 10 / 40 = 25%.
The numerator stays 10 because it counts the overlap. The denominators differ because each question starts with a different group. This matches the conditional-probability rule: divide the overlap by the probability or count of the event after the bar [S1].
What’s a quick way to avoid swapping them when I solve another problem?
Translate the condition into a group before calculating. “Given B” means you have narrowed the cases down to B, so use the B total. “Given A” means use the A total. A reusable sequence is:
- Define A and B in plain language.
- Find the cases in both A and B. That is the numerator in either direction.
- For P(A|B), divide by the B total.
- For P(B|A), divide by the A total.
- Check that the denominator matches the event after the bar.
Here, “given chess player” selects the B total, 20; “given comic reader” selects the A total, 40. The results describe only the constructed counts in this example.
A practical checklist
Write down what A and B mean before using symbols.
Mark the overlap: the count that satisfies both conditions.
For P(A|B), use the total number of B cases as the denominator.
For P(B|A), use the total number of A cases as the denominator.
Check that the denominator matches the condition after the bar.
A few more useful answers
Can the two reverse conditional probabilities ever be equal?
Yes. They are equal when the overlap divided by the B total equals the overlap divided by the A total. They are not automatically equal; compare the counts in the specific problem.
Does P(A|B) mean the probability of A and B happening together?
No. P(A|B) is the proportion of B cases that are also A. The probability of both together is a different quantity, written P(A and B).
Do these percentages describe a real population?
No. The counts here are constructed to demonstrate the calculation. For a particular problem, use its own counts.
Sources
References used for this guide.
- [S1] OpenStax Introductory Statistics 2e, Section 3.3: Two Basic Rules of Probability openstax.org · Checked