IB Math AI Topic 4.6: Venn diagrams, trees and combined events | Notes & Flashcards

Key concepts in Venn diagrams, trees and combined events

Key Idea: Topic 4.6 is about two or more events. Draw them first: a Venn diagram for groups that overlap, a tree diagram for one thing after another. Then use the rules: addition, mutually exclusive, independent, conditional.

Venn diagrams

A ∩ B (intersection): The overlap of the two circles — elements in both A and B. Fill this region first when given n(A∩B).
A ∪ B (union): Everything in either circle (or both). n(A∪B) = n(A) + n(B) − n(A∩B).
A′ (complement): Everything outside circle A: B only and the outside of both circles.
Only in A: n(A) − n(A∩B). The part of A not shared with B. Fill this after the intersection.
Outside both circles: n(U) − n(A∪B). Elements in neither set.

Tree diagrams

With replacement: After the first draw, the item is put back. All probabilities on the second set of branches are the same as the first.
Without replacement: After the first draw, the item is removed. The denominator decreases by 1 on second branches, and the numerator changes depending on what was drawn first.
Reading the tree: Multiply along branches (AND). Add across outcomes that match (OR). Final probabilities in each path must sum to 1.

Rules

Rule	Formula	When to use
Addition	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	A or B: subtract the overlap once.
Mutually exclusive	P(A ∩ B) = 0	They cannot happen together.
Independent	P(A ∩ B) = P(A) × P(B)	One does not change the probability of the other.
Conditional	P(B \| A) = P(A ∩ B) ÷ P(A)	"Given A": look only at the A group.
At least one	1 − P(none)	Quicker than adding every case.

When filling a Venn diagram: always start with the intersection (A∩B), then fill in 'only A' and 'only B', then 'neither'. This order prevents double-counting. All probabilities in a tree diagram (summing across all final branches) must add to 1. Use this as a check.

Paper 1: You may be given a partially completed Venn diagram and asked to find a probability. Read n(A∩B) from the overlap region and use the conditional probability formula. Paper 2: Multi-stage probability problems almost always benefit from a tree diagram. Show all branches with their probabilities, then highlight the relevant paths.

IB-style question [7 marks]

In a group of 90 students, 40 like Mathematics, 30 like Physics and 12 like both subjects. One student is chosen at random. (a) Find the probability that the student likes Mathematics or Physics (or both).

Step by step:

Write each basic probability as the count over the total of 90.
$P(M) = \frac{40}{90},\quad P(Ph) = \frac{30}{90},\quad P(M\cap Ph) = \frac{12}{90}$
Use the addition rule and subtract the overlap so it is not counted twice.
$P(M\cup Ph) = P(M) + P(Ph) - P(M\cap Ph)$
Substitute the values.
$= \frac{40}{90} + \frac{30}{90} - \frac{12}{90} = \frac{58}{90} = \frac{29}{45} \approx 0.644$

Final answer:

29/45 ≈ 0.644.

Part (b)

(b) Given that the student likes Mathematics, find the probability that the student also likes Physics.

Step by step:

Use conditional probability: divide the overlap by the probability of the condition.
$P(Ph\mid M) = \frac{P(M\cap Ph)}{P(M)} = \frac{12/90}{40/90}$
Simplify.
$= \frac{12}{40} = \frac{3}{10} = 0.3$

Final answer:

0.3.

Part (c)

Step by step:

Events are independent only if P(M)×P(Ph) equals P(M∩Ph). Compare the two.
$P(M)\times P(Ph) = \frac{40}{90}\times\frac{30}{90} = \frac{4}{27} \approx 0.148$
Since 0.148 ≠ 12/90 ≈ 0.133, the product does not match the overlap.
$\frac{4}{27} \ne \frac{12}{90} \;\Rightarrow\; \text{not independent}$

Final answer:

Not independent (0.148 ≠ 0.133).

IB-style question [6 marks]

An email filter classifies messages. 30% of all emails are spam. Of the spam emails, 90% contain the word "free". Of the non-spam emails, only 5% contain the word "free". (a) Find the probability that a randomly chosen email contains the word "free".

Step by step:

An email can contain 'free' along two tree paths: spam-and-free, or not-spam-and-free. Find each path, then add.
$P(\text{free}) = P(\text{spam})P(\text{free}\mid \text{spam}) + P(\text{not spam})P(\text{free}\mid \text{not spam})$
Substitute the branch probabilities.
$P(\text{free}) = 0.30 \times 0.90 + 0.70 \times 0.05 = 0.27 + 0.035 = 0.305$

Final answer:

P(free) = 0.305.

Part (b)

(b) An email is found to contain the word "free". Find the probability that it is spam.

Step by step:

This is the reverse direction — use the conditional probability rule. The intersection 'spam and free' is the spam path from part (a).
$P(\text{spam}\mid \text{free}) = \frac{P(\text{spam} \cap \text{free})}{P(\text{free})} = \frac{0.27}{0.305}$
Evaluate.
$P(\text{spam}\mid \text{free}) = 0.885\ \text{(3 s.f.)}$

Final answer:

P(spam | free) = 0.27/0.305 = 0.885 (3 s.f.).

Key concepts in Venn diagrams, trees and combined events

Key Idea: Topic 4.6 is about two or more events. Draw them first: a Venn diagram for groups that overlap, a tree diagram for one thing after another. Then use the rules: addition, mutually exclusive, independent, conditional.

Venn diagrams

A ∩ B (intersection): The overlap of the two circles — elements in both A and B. Fill this region first when given n(A∩B).
A ∪ B (union): Everything in either circle (or both). n(A∪B) = n(A) + n(B) − n(A∩B).
A′ (complement): Everything outside circle A: B only and the outside of both circles.
Only in A: n(A) − n(A∩B). The part of A not shared with B. Fill this after the intersection.
Outside both circles: n(U) − n(A∪B). Elements in neither set.

Tree diagrams

With replacement: After the first draw, the item is put back. All probabilities on the second set of branches are the same as the first.
Without replacement: After the first draw, the item is removed. The denominator decreases by 1 on second branches, and the numerator changes depending on what was drawn first.
Reading the tree: Multiply along branches (AND). Add across outcomes that match (OR). Final probabilities in each path must sum to 1.

Rules

Rule	Formula	When to use
Addition	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	A or B: subtract the overlap once.
Mutually exclusive	P(A ∩ B) = 0	They cannot happen together.
Independent	P(A ∩ B) = P(A) × P(B)	One does not change the probability of the other.
Conditional	P(B \| A) = P(A ∩ B) ÷ P(A)	"Given A": look only at the A group.
At least one	1 − P(none)	Quicker than adding every case.

When filling a Venn diagram: always start with the intersection (A∩B), then fill in 'only A' and 'only B', then 'neither'. This order prevents double-counting. All probabilities in a tree diagram (summing across all final branches) must add to 1. Use this as a check.

Paper 1: You may be given a partially completed Venn diagram and asked to find a probability. Read n(A∩B) from the overlap region and use the conditional probability formula. Paper 2: Multi-stage probability problems almost always benefit from a tree diagram. Show all branches with their probabilities, then highlight the relevant paths.

IB-style question [7 marks]

Step by step:

Write each basic probability as the count over the total of 90.
$P(M) = \frac{40}{90},\quad P(Ph) = \frac{30}{90},\quad P(M\cap Ph) = \frac{12}{90}$
Use the addition rule and subtract the overlap so it is not counted twice.
$P(M\cup Ph) = P(M) + P(Ph) - P(M\cap Ph)$
Substitute the values.
$= \frac{40}{90} + \frac{30}{90} - \frac{12}{90} = \frac{58}{90} = \frac{29}{45} \approx 0.644$

Final answer:

29/45 ≈ 0.644.

Part (b)

(b) Given that the student likes Mathematics, find the probability that the student also likes Physics.

Step by step:

Use conditional probability: divide the overlap by the probability of the condition.
$P(Ph\mid M) = \frac{P(M\cap Ph)}{P(M)} = \frac{12/90}{40/90}$
Simplify.
$= \frac{12}{40} = \frac{3}{10} = 0.3$

Final answer:

0.3.

Part (c)

Step by step:

Events are independent only if P(M)×P(Ph) equals P(M∩Ph). Compare the two.
$P(M)\times P(Ph) = \frac{40}{90}\times\frac{30}{90} = \frac{4}{27} \approx 0.148$
Since 0.148 ≠ 12/90 ≈ 0.133, the product does not match the overlap.
$\frac{4}{27} \ne \frac{12}{90} \;\Rightarrow\; \text{not independent}$

Final answer:

Not independent (0.148 ≠ 0.133).

IB-style question [6 marks]

Step by step:

An email can contain 'free' along two tree paths: spam-and-free, or not-spam-and-free. Find each path, then add.
$P(\text{free}) = P(\text{spam})P(\text{free}\mid \text{spam}) + P(\text{not spam})P(\text{free}\mid \text{not spam})$
Substitute the branch probabilities.
$P(\text{free}) = 0.30 \times 0.90 + 0.70 \times 0.05 = 0.27 + 0.035 = 0.305$

Final answer:

P(free) = 0.305.

Part (b)

(b) An email is found to contain the word "free". Find the probability that it is spam.

Step by step:

This is the reverse direction — use the conditional probability rule. The intersection 'spam and free' is the spam path from part (a).
$P(\text{spam}\mid \text{free}) = \frac{P(\text{spam} \cap \text{free})}{P(\text{free})} = \frac{0.27}{0.305}$
Evaluate.
$P(\text{spam}\mid \text{free}) = 0.885\ \text{(3 s.f.)}$

Final answer:

P(spam | free) = 0.27/0.305 = 0.885 (3 s.f.).

IB Math AI — Venn diagrams, trees and combined events

Key concepts in Venn diagrams, trees and combined events

Venn diagrams

Tree diagrams

Rules

IB-style question [7 marks]

Part (b)

Part (c)

IB-style question [6 marks]

Part (b)

What you'll learn in Topic 4.6

Study resources — 4.6 Venn diagrams, trees and combined events

Venn diagrams

Tree diagrams

Combined events and independence

Counting combined outcomes

Conditional probability

Ready to study Venn diagrams, trees and combined events?

Ready to practice?

IB Math AI — Venn diagrams, trees and combined events

Key concepts in Venn diagrams, trees and combined events

Venn diagrams

Tree diagrams

Rules

IB-style question [7 marks]

Part (b)

Part (c)

IB-style question [6 marks]

Part (b)

What you'll learn in Topic 4.6

Study resources — 4.6 Venn diagrams, trees and combined events

Venn diagrams

Tree diagrams

Combined events and independence

Counting combined outcomes

Conditional probability

Ready to study Venn diagrams, trees and combined events?

Ready to practice?