--- title: "Conditional Probability from Tables and Venn Diagrams" description: "Work out the probability of one event given that another has happened, by restricting attention to the row, column or region the condition allows." canonical: https://lightmysky.com/learn/mathematics/conditional-probability-from-tables-and-venn-diagrams-mt_Vx88ugViYP source: https://lightmysky.com/learn/mathematics/conditional-probability-from-tables-and-venn-diagrams-mt_Vx88ugViYP.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Conditional Probability from Tables and Venn Diagrams Work out the probability of one event given that another has happened, by restricting attention to the row, column or region the condition allows. Subject: Mathematics · Area: Probability · Ages 15 to 16 Page: https://lightmysky.com/learn/mathematics/conditional-probability-from-tables-and-venn-diagrams-mt_Vx88ugViYP ## Ready when they can - Read a conditional probability from a two-way table using the total for the given group as the denominator - Find a conditional probability from a region of a Venn diagram - Explain in words how P(A given B) differs from P(B given A) ## Lesson: Given that changes everything You have used totals in two-way tables. A conditional probability keeps only a given group. P(A given B), also written P(A|B), asks how likely A is among B cases. For uniform random selection, divide the count in both A and B by the count in B. Every case is equally likely to be selected. The given group must contain cases: division by zero is undefined. **Example.** Choose a pupil uniformly at random. Group Tennis No tennis Total Girls 30 90 120 Boys 20 60 80 Total 50 150 200 Given girl, keep the girls row: P(tennis given girl) = 30/120 = 0.25. Given tennis player, keep the tennis column: P(girl given tennis player) = 30/50 = 0.6. Both calculations count the same 30 pupils in the numerator. A Venn diagram separates A only, both, B only and neither. Suppose A only has 12 cases, both has 8 and B only has 20. For uniform selection, P(A given B) uses all of B: 8 + 20 = 28 cases. Only the 8 overlap cases are also in A, so the probability is 8/28 = 2/7. Reversing the condition uses all of A: P(B given A) = 8/(12 + 8) = 0.4. Cases in neither set are excluded from both denominators. *(drawing: These bars show the two separate counts inside B, not a Venn diagram. Add them for B's total, then use the overlap over that total.)* Reversing the condition keeps the overlap but changes the given group. The answers can differ or be equal. Equal group totals give equal answers; an empty overlap gives zero in both directions when both groups have cases. You can recover an overlap count by multiplying a conditional probability by its given total. If B has 25 cases and P(A given B) = 0.2, the overlap contains 0.2 × 25 = 5 cases. Divide that count by A's total for the reverse probability. Complete missing table counts before dividing. Add cells to find a row total, or subtract known cells from that total to find a missing cell. For P(not A given B), subtract the overlap from B's total to count favourable cases. A condition can allow several rows, such as Year 10 or Year 11. Add their row totals for the denominator and their favourable cells for the numerator. **Example.** Some tables contain whole-group probabilities instead of counts. Divide the joint cell for both events by the total for the given event. If the probability of both A and B is 0.12 and B has probability 0.4, then P(A given B) = 0.12/0.4 = 0.3. Both entries must use the same whole-group scale. A test's detection rate is P(positive given condition). It describes cases with the condition. P(condition given positive) describes the positive results instead. Knowing the detection rate and that the condition is rare does not determine the reverse probability. You also need information about positives among cases without the condition. If none of those cases test positive, every positive comes from a case with the condition. **Example.** In a made-up inspection, 10 faulty parts and 40 sound parts test positive. Choose a part uniformly at random. Given positive, the group has 10 + 40 = 50 parts. P(faulty given positive) = 10/50 = 0.2. Here most positives are false alarms: the 40 sound parts outnumber the 10 faulty parts in the positive group. **Recap.** Circle the given group first, then divide the overlap by that group alone. *Lesson, questions and flashcards written for this dataset. Inspection examples use hypothetical data.* ## Practice 17 questions on this page, each with its working shown. ## Needs first - [Sets & Venn Diagrams](https://lightmysky.com/learn/mathematics/sets-and-venn-diagrams-mt_1YwOCMMwD8) - [Tree Diagrams Without Replacement](https://lightmysky.com/learn/mathematics/tree-diagrams-without-replacement-mt_NfiAY_wMwZ) - [Two-Way Tables](https://lightmysky.com/learn/mathematics/two-way-tables-mt_oD1Rgg52n3) ## Opens up - [Setting Up a Hypothesis Test](https://lightmysky.com/learn/mathematics/setting-up-a-hypothesis-test-mt_JmpLNHK0Nf) - [Conditional Probability, Independence and Bayes' Theorem](https://lightmysky.com/learn/mathematics/conditional-probability-independence-and-bayes-theorem-mt_wIi_XRzwwK)