IB Math AA SL Topic 4.6: Combined & conditional events | Notes & Flashcards | Aimnova

IB Math AA SL — Combined & conditional events

Topic 4.6 of IB Mathematics: Analysis and Approaches covers Combined & conditional events, which is part of Unit 4: Statistics & Probability. Students explore key concepts including Venn diagrams, Tree diagrams, Independent events. A strong understanding of combined & conditional events is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Combined & conditional events

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

P(A\cup B) = P(A) + P(B) - P(A\cap B)

$P(A\cup B)$: probability of A or B (at least one)
$P(A\cap B)$: probability of both — the overlap you subtract

P(A\cap B) = P(A)\times P(B)\quad(\text{independent})

$\text{independent}$: 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.
$24-10 = 14,\quad 18-10 = 8$
Neither = total − (only S + both + only F).
$40-(14+10+8) = 8$
Each probability is region ÷ total.
$P(\text{only S}) = \tfrac{14}{40},\ P(\text{neither}) = \tfrac{8}{40}$

Final answer:

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.
$\tfrac{5}{8}\cdot\tfrac{3}{7} + \tfrac{3}{8}\cdot\tfrac{5}{7}$
Totals drop on the second draw (7 left).
$= \tfrac{15}{56} + \tfrac{15}{56}$
Add the paths.
$= \tfrac{30}{56} = \tfrac{15}{28}$

Final answer:

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.
$0.7 = 0.4 + P(B) - 0.4\,P(B)$
Collect the P(B) terms.
$0.3 = 0.6\,P(B)$
Solve the linear equation.
$P(B) = 0.5$

Final answer:

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.

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)).

IB Math AA SL — Combined & conditional events

Key concepts in Combined & conditional events

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

P(A\cup B) = P(A) + P(B) - P(A\cap B)

$P(A\cup B)$: probability of A or B (at least one)
$P(A\cap B)$: probability of both — the overlap you subtract

P(A\cap B) = P(A)\times P(B)\quad(\text{independent})

$\text{independent}$: 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.
$24-10 = 14,\quad 18-10 = 8$
Neither = total − (only S + both + only F).
$40-(14+10+8) = 8$
Each probability is region ÷ total.
$P(\text{only S}) = \tfrac{14}{40},\ P(\text{neither}) = \tfrac{8}{40}$

Final answer:

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.
$\tfrac{5}{8}\cdot\tfrac{3}{7} + \tfrac{3}{8}\cdot\tfrac{5}{7}$
Totals drop on the second draw (7 left).
$= \tfrac{15}{56} + \tfrac{15}{56}$
Add the paths.
$= \tfrac{30}{56} = \tfrac{15}{28}$

Final answer:

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.
$0.7 = 0.4 + P(B) - 0.4\,P(B)$
Collect the P(B) terms.
$0.3 = 0.6\,P(B)$
Solve the linear equation.
$P(B) = 0.5$

Final answer:

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.

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)).

IB Math AA SL — Combined & conditional events

Key concepts in Combined & conditional events

🔵 Sets, tools & the core rules

✏️ IB-style worked examples

IB-style question — fill a Venn and read a probability

IB-style question — tree diagram without replacement

IB-style question — independence inside the addition rule

Exam Tips

What you'll learn in Topic 4.6

Study resources — 4.6 Combined & conditional events

Venn diagrams

Tree diagrams

Independent events

Ready to study Combined & conditional events?

Ready to practice?

IB Math AA SL — Combined & conditional events

Key concepts in Combined & conditional events

🔵 Sets, tools & the core rules

✏️ IB-style worked examples

IB-style question — fill a Venn and read a probability

IB-style question — tree diagram without replacement

IB-style question — independence inside the addition rule

Exam Tips

What you'll learn in Topic 4.6

Study resources — 4.6 Combined & conditional events

Venn diagrams

Tree diagrams

Independent events

Ready to study Combined & conditional events?

Ready to practice?