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NotesMath AI HLTopic 4.6
Unit 4 · Statistics & Probability · Topic 4.6

IB Math AI HL — Venn diagrams

IB Mathematics AI SL topic covering core concepts and exam-style applications.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Exam technique guidePractice questions

Key concepts in Venn diagrams

Key Idea: Topic 4.6 builds on the probability rules from 4.5 by giving you two powerful visual tools: Venn diagrams (for showing how sets overlap and how probabilities relate) and tree diagrams (for sequential events where the outcome of the first affects the second). Both tools organise information — the key is knowing when to use each one.

✅ Venn diagrams

A ∩ B (intersection)
The overlap of the two circles — elements in both A and B. Fill this region first when given n(A∩B).
A ∪ B (union)
Everything in either circle (or both). n(A∪B) = n(A) + n(B) − n(A∩B).
A' (complement)
Everything outside circle A — neither in A nor in A∩B.
Only in A
n(A) − n(A∩B). The part of A not shared with B. Fill this after the intersection.
Outside both circles
n(S) − n(A∪B). Elements in neither set.

✅ Tree diagrams for conditional probability

With replacement
After the first draw, the item is put back. All probabilities on the second set of branches are the same as the first.
Without replacement
After the first draw, the item is removed. The denominator decreases by 1 on second branches, and the numerator changes depending on what was drawn first.
Reading the tree
Multiply along branches (AND). Add across outcomes that match (OR). Final probabilities in each path must sum to 1.
Example: Venn diagram: In a class of 30: 18 study French (F), 12 study Spanish (S), 7 study both. n(F∩S) = 7. n(F only) = 18−7 = 11. n(S only) = 12−7 = 5. n(neither) = 30−(11+7+5) = 7. P(F|S) = P(F∩S)/P(S) = (7/30)/(12/30) = 7/12 Tree diagram without replacement: 4 red, 3 blue balls. Draw two without replacement. P(red then blue) = (4/7) × (3/6) = 12/42 = 2/7
When filling a Venn diagram: always start with the intersection (A∩B), then fill in 'only A' and 'only B', then 'neither'. This order prevents double-counting. All probabilities in a tree diagram (summing across all final branches) must add to 1. Use this as a check.
Paper 1: You may be given a partially completed Venn diagram and asked to find a probability. Read n(A∩B) from the overlap region and use the conditional probability formula. Paper 2: Multi-stage probability problems almost always benefit from a tree diagram. Show all branches with their probabilities, then highlight the relevant paths.

IB-style question [6 marks]

An email filter classifies messages. 30% of all emails are spam. Of the spam emails, 90% contain the word "free". Of the non-spam emails, only 5% contain the word "free". (a) Find the probability that a randomly chosen email contains the word "free". (b) An email is found to contain the word "free". Find the probability that it is spam.

Step by step:

  1. (a) An email can contain 'free' along two tree paths: spam-and-free, or not-spam-and-free. Find each path, then add.

    P(free)=P(spam)P(free∣spam)+P(not spam)P(free∣not spam)P(\text{free}) = P(\text{spam})P(\text{free}\mid \text{spam}) + P(\text{not spam})P(\text{free}\mid \text{not spam})P(free)=P(spam)P(free∣spam)+P(not spam)P(free∣not spam)
  2. Substitute the branch probabilities.

    P(free)=0.30×0.90+0.70×0.05=0.27+0.035=0.305P(\text{free}) = 0.30 \times 0.90 + 0.70 \times 0.05 = 0.27 + 0.035 = 0.305P(free)=0.30×0.90+0.70×0.05=0.27+0.035=0.305
  3. (b) This is the reverse direction — use the conditional probability rule. The intersection 'spam and free' is the spam path from part (a).

    P(spam∣free)=P(spam∩free)P(free)=0.270.305P(\text{spam}\mid \text{free}) = \frac{P(\text{spam} \cap \text{free})}{P(\text{free})} = \frac{0.27}{0.305}P(spam∣free)=P(free)P(spam∩free)​=0.3050.27​
  4. Evaluate.

    P(spam∣free)=0.885 (3 s.f.)P(\text{spam}\mid \text{free}) = 0.885\ \text{(3 s.f.)}P(spam∣free)=0.885 (3 s.f.)
Final answer:

(a) P(free) = 0.305. (b) P(spam | free) = 0.27/0.305 = 0.885 (3 s.f.).

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What you'll learn in Topic 4.6

  • 4.6.1 Venn Diagrams
  • 4.6.2 Tree Diagrams and Conditional Probability
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 4.6 Venn diagrams

4.6.1

Venn Diagrams

Notes
4.6.2

Tree Diagrams and Conditional Probability

Notes

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Topic 4.6 Venn diagrams forms a core part of Unit 4: Statistics & Probability in IB Math AI HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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