IB Math AA SL Topic 4.11: Conditional probability | Notes & Flashcards | Aimnova

IB Math AA SL — Conditional probability

Topic 4.11 of IB Mathematics: Analysis and Approaches covers Conditional probability, which is part of Unit 4: Statistics & Probability. Students explore key concepts including Conditional probability. A strong understanding of conditional probability is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Conditional probability

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

P(A\mid B) = \frac{P(A\cap B)}{P(B)}

$A\mid B$: A given that B has happened (read '|' as 'given')
$P(A\cap B)$: probability of BOTH A and B
$P(B)$: 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

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.
$\text{boys} = 40 + 25 = 65$
Of those boys, count the ones who cycle, then divide.
$P(\text{cycles}\mid\text{boy}) = \frac{40}{65} = \frac{8}{13}$

Final answer:

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.
$P(A\cap B) = P(A) + P(B) - P(A\cup B) = 0.5 + 0.4 - 0.7 = 0.2$
Apply the conditional formula — divide by the given event B.
$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5$

Final answer:

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'.
$0.6(0.03) + 0.4(0.08) = 0.018 + 0.032 = 0.05$
Conditional: the matching path (Y and faulty) over the total.
$P(Y\mid\text{faulty}) = \frac{0.4(0.08)}{0.05} = \frac{0.032}{0.05} = 0.64$

Final answer:

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.

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).

IB Math AA SL — Conditional probability

Key concepts in Conditional probability

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

P(A\mid B) = \frac{P(A\cap B)}{P(B)}

$A\mid B$: A given that B has happened (read '|' as 'given')
$P(A\cap B)$: probability of BOTH A and B
$P(B)$: 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

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

Step by step:

'Given boy' restricts the sample space to the 65 boys — that is the denominator.
$\text{boys} = 40 + 25 = 65$
Of those boys, count the ones who cycle, then divide.
$P(\text{cycles}\mid\text{boy}) = \frac{40}{65} = \frac{8}{13}$

Final answer:

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.
$P(A\cap B) = P(A) + P(B) - P(A\cup B) = 0.5 + 0.4 - 0.7 = 0.2$
Apply the conditional formula — divide by the given event B.
$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5$

Final answer:

P(A | B) = 0.5.

IB-style question — reverse a condition on a tree

Step by step:

Total P(faulty) — add the two paths that end in 'faulty'.
$0.6(0.03) + 0.4(0.08) = 0.018 + 0.032 = 0.05$
Conditional: the matching path (Y and faulty) over the total.
$P(Y\mid\text{faulty}) = \frac{0.4(0.08)}{0.05} = \frac{0.032}{0.05} = 0.64$

Final answer:

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.

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).

IB Math AA SL — Conditional probability

Key concepts in Conditional probability

🎯 The conditional formula

🧭 Where the pieces come from

✏️ IB-style worked examples

IB-style question — read a conditional from a two-way table

IB-style question — find the intersection first (Paper 1)

IB-style question — reverse a condition on a tree

Exam Tips

What you'll learn in Topic 4.11

Study resources — 4.11 Conditional probability

Conditional probability

Ready to study Conditional probability?

Ready to practice?

IB Math AA SL — Conditional probability

Key concepts in Conditional probability

🎯 The conditional formula

🧭 Where the pieces come from

✏️ IB-style worked examples

IB-style question — read a conditional from a two-way table

IB-style question — find the intersection first (Paper 1)

IB-style question — reverse a condition on a tree

Exam Tips

What you'll learn in Topic 4.11

Study resources — 4.11 Conditional probability

Conditional probability

Ready to study Conditional probability?

Ready to practice?