Key Idea: Conditional probability is about updating a probability once you know something has already happened — given that B occurred, how likely is A? It shows up as 'given that…' questions, often from a two-way table, Venn or tree, on both papers.
🎯 The conditional formula
- A given that B has happened (read '|' as 'given')
- probability of BOTH A and B
- the GIVEN event — the new, reduced sample space you divide by
Knowing B happened throws away every outcome where B did not — so you only look inside B, which is why you divide by P(B), the event after the '|'. Note A|B and B|A differ: divide by whichever event is given.
🧭 Where the pieces come from
| You're given… | How to get the answer |
|---|---|
| P(A∩B) and P(B) directly | Straight into the formula: P(A∩B) ÷ P(B). |
| P(A), P(B), P(A∪B) | First recover the intersection: P(A∩B) = P(A) + P(B) − P(A∪B), then divide. |
| A two-way table or Venn | Restrict to the given group: count(both) ÷ count(given group) — only look in that row/region. |
| A tree diagram | A 2nd-stage branch is already conditional. To reverse it: matching path ÷ total of all paths to that outcome. |
A and B are independent when knowing one tells you nothing about the other: P(A | B) = P(A) (equivalently P(A∩B) = P(A) × P(B)). If P(A | B) ≠ P(A), they are dependent.
✏️ IB-style worked examples
IB-style question — read a conditional from a two-way table
120 students are asked if they cycle to school: Girls: 30 cycle, 25 do not. Boys: 40 cycle, 25 do not. A student chosen at random is a boy. Find the probability that he cycles, P(cycles | boy).
Step by step:
'Given boy' restricts the sample space to the 65 boys — that is the denominator.
Of those boys, count the ones who cycle, then divide.
P(cycles | boy) = 8/13 ≈ 0.615.
IB-style question — find the intersection first (Paper 1)
For events A and B, P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find P(A | B) without a calculator.
Step by step:
Recover the intersection from the addition rule.
Apply the conditional formula — divide by the given event B.
P(A | B) = 0.5.
IB-style question — reverse a condition on a tree
A factory uses machine X for 60% of parts and machine Y for 40%. X produces a faulty part 3% of the time; Y, 8%. A part is found to be faulty. Find the probability it came from machine Y, P(Y | faulty).
Step by step:
Total P(faulty) — add the two paths that end in 'faulty'.
Conditional: the matching path (Y and faulty) over the total.
P(Y | faulty) = 0.64.
Important: You must divide by P(B), the given event — not the whole sample space and not P(A). 'Given boy' → divide by total boys, not all students. On a tree, the denominator is the total probability of the given outcome (all paths), not the single path.
Tap each card to reveal the answer.
P(A∩B) = 0.18, P(B) = 0.3. Find P(A | B). 0.6 — divide 0.18 by the given event P(B) = 0.3.
P(A) = 0.6, P(B) = 0.5, P(A∪B) = 0.9. Find P(A∩B). 0.2 — addition rule: 0.6 + 0.5 − 0.9.
Of 50 women, 35 own a phone. Find P(phone | woman). 35/50 = 7/10 = 0.7 — restrict to the 50 women.
P(A | B) = 0.4 and P(A) = 0.4. Are A and B independent? Yes — P(A | B) = P(A), so B tells you nothing about A.
On a tree, P(rain∩late) = 0.12 and P(late) = 0.3. Find P(rain | late). 0.4 — matching path 0.12 over the total P(late) = 0.3.
Exam tips
- Divide by the event AFTER the '|' — the given event is your denominator.
- From a table or Venn, restrict to the given row/region: count(both) ÷ count(given group).
- Only have P(A), P(B), P(A∪B)? Find P(A∩B) = P(A) + P(B) − P(A∪B) first.
- On a tree, reverse a condition with: matching path ÷ total of ALL paths to that outcome.
- Independent ⇔ P(A | B) = P(A) ⇔ P(A∩B) = P(A) × P(B).