Venn and Tree Diagrams (AA HL)
Most probability questions boil down to picking the right diagram. A Venn diagram sorts events by overlap and is the natural home for the addition rule; a tree diagram lays out a sequence of choices branch by branch and is the natural home for multiplying along a path. This page covers combining events with both tools, including what changes when a draw is made without replacement. It's part of the broader Probability topic.
22 questions on this sub-topic.
Combining events
Covered under IB syllabus reference SL4.6: Venn diagrams, tree diagrams, sample space diagrams and tables of outcomes, and the rule for combined events \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), with \(P(A\cap B)=0\) when the events are mutually exclusive.
Addition rule (Venn)
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Subtract the overlap once so it isn't counted twice. If the events are mutually exclusive, the overlap is \(0\) and the rule reduces to \(P(A)+P(B)\).
Finding \(P(A)\) with a tree
When \(P(A)\) isn't given directly, build it from a tree diagram: sum \(P(B_i)P(A\mid B_i)\) over every branch that leads to \(A\), for up to three events \(B_1,B_2,B_3\).
Need the full syllabus wording and formula-booklet reference table? See Probability.
Worked examples
Of 100 students, 60 study Spanish, S, and 25 of those also study French, F. Altogether 40 study French.
(a) Find \(P(F)\).
(b) Find \(P(F\mid S)\).
Worked solution
(a) \(P(F)\). \(P(F)=\dfrac{40}{100}=0.4.\) A1
(b) \(P(F\mid S)\). Condition on the 60 Spanish students; of these, 25 also study French: \(P(F\mid S)=\dfrac{25}{60}\approx 0.417.\) M1
A1
A box has 12 items, 3 defective. Two are drawn without replacement. Given the first is defective, find \(P(\text{second is defective}).\)
Worked solution
After one defective is removed, 11 items remain with 2 defective. M1
\(P(\text{2nd defective}) = \dfrac{2}{11} \approx 0.182.\) A1
Two fair dice are rolled and the scores added.
(a) State the number of outcomes.
(b) Find the probability the total is 7.
Worked solution
(a) Number of outcomes. Each die has 6 faces and they are independent: \(6\times6=36.\) A1
(b) Total is 7. The favourable pairs are \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) - 6 ways. M1
So \(P=\dfrac{6}{36}=\dfrac16.\) A1
Common mistakes
- Forgetting "without replacement" changes the second probability. If a card, ball or item isn't put back, the total and the count of favourable outcomes both shrink for the next draw - each branch of the tree needs a fresh denominator.
- Double-subtracting the overlap in a Venn diagram. The region where both events happen is drawn once but represents \(P(A\cap B)\) - label it directly, then work outward, rather than subtracting it twice by accident.
- Forgetting to branch on complements in a tree diagram. Every stage needs both outcomes drawn, since "not \(A\)" branches often carry the probabilities that matter most for the next stage.
Ready to practise properly?
22 Venn and tree diagram questions, marked instantly like the real exam.
Quick answers
What is the formula for the union of two events?
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). For mutually exclusive events \(P(A\cap B)=0\), so it simplifies to \(P(A)+P(B)\).
How do you find a probability from a tree diagram?
Multiply along each branch to get the probability of that path, then add the results for every path that leads to the outcome you want. For GDC guidance on organising these calculations, see the Probability GDC section.