Key Idea: This topic is about combining the chances of two events — A and B, A or B — using Venn diagrams, tree diagrams and the addition rule. It runs through both papers, non-calculator and GDC alike.
🔵 Sets, tools & the core rules
| Symbol / tool | Read it as | Key move |
|---|---|---|
| A ∩ B | A and B — the overlap | multiply for 'both' |
| A ∪ B | A or B — at least one | addition rule (subtract overlap) |
| A′ | not A — outside A | P(A′) = 1 − P(A) |
| Venn diagram | regions of a count/probability | fill overlap first; prob = region ÷ total |
| Tree diagram | branches per stage (each sums to 1) | multiply along a path, add paths |
| Without replacement | item not put back | reduce type and total by 1 |
- probability of A or B (at least one)
- probability of both — the overlap you subtract
- one event doesn't change the other's probability
Mutually exclusive: can't both happen. P(A ∩ B) = 0. P(A ∪ B) = P(A) + P(B).
Independent: one doesn't affect the other. P(A ∩ B) = P(A) × P(B). usually a non-zero overlap.
✏️ IB-style worked examples
IB-style question — fill a Venn and read a probability
In a class of 40 students, 24 study Spanish (S), 18 study French (F) and 10 study both. Find P(studies only Spanish) and P(studies neither).
Step by step:
Overlap is 10. Subtract for the 'only' regions.
Neither = total − (only S + both + only F).
Each probability is region ÷ total.
P(only Spanish) = 14/40 = 7/20; P(neither) = 8/40 = 1/5.
IB-style question — tree diagram without replacement
A box has 5 blue and 3 green counters. Two are taken out without replacement. Find the probability of getting one of each colour.
Step by step:
Two paths match: blue-then-green and green-then-blue.
Totals drop on the second draw (7 left).
Add the paths.
P(one of each) = 30/56 = 15/28.
IB-style question — independence inside the addition rule
Events A and B are independent, with P(A) = 0.4 and P(A ∪ B) = 0.7. Find P(B).
Step by step:
Independent, so P(A ∩ B) = P(A)·P(B). Sub into the addition rule.
Collect the P(B) terms.
Solve the linear equation.
P(B) = 0.5.
Important: They are different. Mutually exclusive → P(A ∩ B) = 0. Independent → P(A ∩ B) = P(A)·P(B) (non-zero). Don't drop the overlap in the addition rule unless you're told the events can't both happen.
Tap each card to reveal the answer.
P(A) = 0.5, P(B) = 0.3, P(A ∩ B) = 0.1 — find P(A ∪ B) 0.5 + 0.3 − 0.1 = 0.7 — addition rule, subtract the overlap.
Out of 50 people, 30 like tea, 20 like coffee, 12 like both — how many like neither? Like at least one = 30 + 20 − 12 = 38, so neither = 50 − 38 = 12.
Bag of 4 red, 6 blue; draw two WITH replacement — P(both red)? Same each draw: 4/10 × 4/10 = 16/100 = 4/25.
Same bag, WITHOUT replacement — P(both red)? Totals drop: 4/10 × 3/9 = 12/90 = 2/15.
P(A) = 0.6, P(B) = 0.5 — are they independent if P(A ∩ B) = 0.3? 0.6 × 0.5 = 0.3 = P(A ∩ B), so yes, independent.
A and B are mutually exclusive, P(A) = 0.35, P(B) = 0.4 — find P(A ∪ B) P(A ∩ B) = 0, so 0.35 + 0.4 = 0.75.
Exam Tips
- Read ∩ as 'and' (overlap), ∪ as 'or' (at least one), ′ as 'not'.
- Venn: fill the overlap first, subtract for 'only' regions, then probability = region ÷ total.
- Tree: multiply along a path, add across matching paths; without replacement the totals drop.
- Use 1 − P(none) for 'at least one' — one product instead of many paths.
- Mutually exclusive (P∩ = 0) is NOT the same as independent (P∩ = P(A)·P(B)).