IB Mathematics · Grade 9

Probability &
Set Theory

20 essential questions to master the most commonly tested concepts. Work through every problem before checking the answer.

20Questions

4Topics

★IB Level

Core Symbols

∪ Union — "OR" (everything in either set)
∩ Intersection — "AND" (only what's in both)
A′ Complement — "NOT A" (everything outside A)
∅ Empty set — has zero elements
⊂ Subset — every element of A is in B
n(A) Number of elements in set A

⚡ Memory Points

🔵 UNION = U-shape = "Use Both"

🟣 INTERSECTION = "IN both"

⬛ COMPLEMENT = "Completely Outside"

🧮 Inclusion-Exclusion:
n(A∪B) = n(A)+n(B)−n(A∩B)

Set Notation · Basic

Let U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {2, 4, 6, 8}, B = {1, 2, 3, 4}.

Find A ∩ B′.

💡 Step order: Find B′ first (elements NOT in B), then intersect with A.

A {6, 8}

B {2, 4}

C {4, 6, 8}

D {1, 3, 5, 6, 7, 8}

Solution

B′ = all elements in U NOT in B = {5, 6, 7, 8}.

A ∩ B′ = elements in both A = {2,4,6,8} and B′ = {5,6,7,8} → {6, 8}.

Common mistake: Students forget to apply complement first and compute A∩B = {2,4} instead.

Set Notation · Inclusion-Exclusion

In a class of 30 students, 18 study French, 14 study Spanish, and 6 study both. How many students study neither French nor Spanish?

💡 Formula: n(F∪S) = n(F) + n(S) − n(F∩S). Then "neither" = Total − n(F∪S).

A 4

B 26

C 6

D 8

Solution

n(F∪S) = 18 + 14 − 6 = 26

Neither = 30 − 26 = 4

Common mistake: Forgetting to subtract the intersection (6) leads to n(F∪S) = 32, which exceeds the total!

Venn Diagram · Reading

From the Venn diagram below, how many elements are in (A ∪ B)′?

💡 (A∪B)′ means everything OUTSIDE both circles — inside U but not in A or B.

A 6

B 4

C 15

D 21

Solution

A∪B contains: 7 (only A) + 3 (both) + 5 (only B) = 15 elements.

Elements outside A∪B (in U only): 4 + 2 = 6.

The numbers outside the circles (4 on the left, 2 on the right) are the elements of (A∪B)′.

Set Operations · Tricky

If n(U) = 40, n(A) = 25, n(B) = 20, and n(A ∩ B) = 10, find n(A′ ∩ B′).

💡 De Morgan's Law: A′ ∩ B′ = (A ∪ B)′. Find n(A∪B) first, then subtract from n(U).

A 15

B 5

C 10

D 20

Solution

n(A∪B) = 25 + 20 − 10 = 35

A′∩B′ = (A∪B)′ → n = 40 − 35 = 5

De Morgan's Law is often tested indirectly — recognize A′∩B′ = (A∪B)′ immediately!

Subsets · Counting

How many subsets does the set S = {a, b, c, d} have?

💡 Rule: A set with n elements has 2ⁿ subsets (including ∅ and S itself).

A 8

B 4

C 16

D 12

Solution

|S| = 4 elements.

Number of subsets = 2⁴ = 16. (This includes ∅ and {a,b,c,d} itself.)

Each element has 2 choices: in or out. So total = 2×2×2×2 = 16.

Key Rules

P(A) = favourable outcomestotal outcomes
0 ≤ P(A) ≤ 1
P(A) + P(A′) = 1

⚡ Memory Points

🎯 CERTAIN event → P = 1

❌ IMPOSSIBLE event → P = 0

🔄 COMPLEMENT: P(not A) = 1 − P(A)

📋 All P must SUM to 1

Basic Probability · Classic

A bag contains 5 red, 3 blue, and 2 green marbles. A marble is picked at random. What is the probability it is NOT red?

💡 P(not red) = 1 − P(red). Or count directly: not-red = blue + green.

A 1/2

B 3/10

C 5/10

D 1/5

Solution

Total marbles = 5 + 3 + 2 = 10.

P(red) = 5/10 = 1/2.

P(NOT red) = 1 − 1/2 = 1/2. (Or: (3+2)/10 = 5/10 = 1/2.)

Sample Space · Dice

Two fair dice are rolled. What is the probability that the sum equals 8?

💡 List all pairs (d₁, d₂) that give sum = 8: (2,6),(3,5),(4,4),(5,3),(6,2). Total outcomes = 6² = 36.

A 1/6

B 5/36

C 1/9

D 7/36

Solution

Total outcomes = 6 × 6 = 36.

Pairs giving sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) — that's 5 pairs.

P = 5/36.

Common mistake: Writing only 3 pairs if (4,4) is counted once instead of recognising symmetry.

Complementary Probability · Tricky

The probability that it rains on any given day is 0.3. What is the probability that it does not rain on either of two consecutive days?

💡 "Not rain" on one day = 0.7. Two independent days: multiply P(no rain) × P(no rain).

A 0.49

B 0.6

C 0.42

D 0.09

Solution

P(no rain one day) = 1 − 0.3 = 0.7.

Days are independent → P(no rain both days) = 0.7 × 0.7 = 0.49.

Watch out: 0.09 = P(rain on both days) = 0.3×0.3. Don't confuse the events!

Addition Rules

General: P(A∪B) = P(A)+P(B)−P(A∩B)

Mutually Exclusive (A∩B=∅):
P(A∪B) = P(A) + P(B)

⚡ Memory Points

🚫 MUTUALLY EXCLUSIVE = can't happen TOGETHER → P(A∩B)=0

🔗 INDEPENDENT = one doesn't AFFECT the other → P(A∩B)=P(A)×P(B)

⚠️ Mutually exclusive ≠ independent!

Mutually Exclusive · Identify

P(A) = 0.4 and P(B) = 0.3. Events A and B are mutually exclusive. Find P(A ∪ B).

💡 Mutually exclusive means they cannot both happen, so P(A∩B) = 0.

A 0.12

B 0.58

C 0.7

D 1.0

Solution

Mutually exclusive → P(A∩B) = 0.

P(A∪B) = 0.4 + 0.3 − 0 = 0.7.

0.12 = P(A)×P(B) which would apply if they were independent — don't mix the two rules!

Independent Events · Test

P(A) = 0.5, P(B) = 0.6, P(A∩B) = 0.3.

Are events A and B independent?

💡 Independence test: Check if P(A∩B) = P(A) × P(B). If equal → independent.

A Yes, because 0.5 + 0.6 = 1.1

B Yes, because 0.5 × 0.6 = 0.3 = P(A∩B)

C No, because P(A∩B) ≠ 0

D No, because P(A∪B) ≠ 1

Solution

Independence test: P(A) × P(B) = 0.5 × 0.6 = 0.3.

P(A∩B) = 0.3 = 0.3 ✓ → Yes, they are independent.

Independent does NOT mean mutually exclusive. Here A∩B can (and does) happen.

Combined Events · Solve for Unknown

P(A) = 0.6, P(A∪B) = 0.8, P(A∩B) = 0.2. Find P(B).

💡 Rearrange: P(B) = P(A∪B) − P(A) + P(A∩B).

A 0.6

B 0.2

C 0.4

D 0.8

Solution

P(A∪B) = P(A) + P(B) − P(A∩B)

0.8 = 0.6 + P(B) − 0.2 → P(B) = 0.8 − 0.6 + 0.2 = 0.4

Conditional Probability

P(A|B) = P(A∩B)P(B)

Read: "Probability of A given B has occurred"

⚡ Memory Points

🌲 TREE: multiply along branches

➕ TREE: add across end branches

📐 P(A|B) = "restrict universe to B"

🔁 WITH replacement → same fraction each time

🚫 WITHOUT replacement → denominator decreases by 1

Conditional Probability · Formula

P(A∩B) = 0.12 and P(B) = 0.4. Find P(A|B).

💡 P(A|B) = P(A∩B) ÷ P(B). Divide, don't subtract!

A 0.28

B 0.48

C 0.3

D 0.052

Solution

P(A|B) = P(A∩B) / P(B) = 0.12 / 0.4 = 0.3.

0.28 = P(B) − P(A∩B), a common but wrong operation. Always divide for conditional probability.

Without Replacement · Tree

A box has 4 red and 3 blue balls. Two balls are drawn without replacement. What is P(both red)?

💡 P(1st red) = 4/7. After removing one red: P(2nd red | 1st red) = 3/6. Multiply.

A 16/49

B 2/7

C 3/7

D 12/42

Solution

P(1st red) = 4/7.

Without replacement: now 3 red out of 6 total → P(2nd red) = 3/6 = 1/2.

P(both red) = (4/7) × (3/6) = 12/42 = 2/7.

16/49 is the answer WITH replacement — notice the denominator stays at 7 in that case.

Tree Diagram · Two-Stage

In a school, 60% of students are female. 30% of female students and 20% of male students play sport. A student is chosen at random. Find the probability the student plays sport.

💡 P(sport) = P(F)×P(S|F) + P(M)×P(S|M). This is the total probability rule.

A 0.26

B 0.50

C 0.18

D 0.30

Solution

P(F) = 0.6, P(M) = 0.4.

P(sport∩F) = 0.6 × 0.3 = 0.18.

P(sport∩M) = 0.4 × 0.2 = 0.08.

P(sport) = 0.18 + 0.08 = 0.26.

Conditional Probability · Reverse

Using Q14's data, given that a student plays sport, what is the probability that the student is female?

💡 P(F|sport) = P(F∩sport) / P(sport). Use your answer from Q14!

A 0.30

B 9/13

C 0.60

D 4/13

Solution

P(F∩sport) = 0.18 (from Q14, step 2).

P(sport) = 0.26 (from Q14).

P(F|sport) = 0.18/0.26 = 18/26 = 9/13 ≈ 0.692.

Venn & Probability · Combined

From the Venn diagram: n(A only) = 8, n(A∩B) = 4, n(B only) = 6, outside both = 2.

A person is chosen at random. Find P(B | A).

💡 P(B|A) = P(A∩B)/P(A). Total in A = "A only" + "A∩B".

A 4/20

B 1/3

C 4/10

D 2/5

Solution

Total = 8+4+6+2 = 20. n(A) = 8+4 = 12. n(A∩B) = 4.

P(B|A) = n(A∩B)/n(A) = 4/12 = 1/3.

Note: 4/10 is wrong — n(B)=10 but we're conditioning on A, not B.

Probability · Tricky Language

A card is drawn from a standard 52-card deck. What is the probability of drawing a King OR a Heart?

💡 Use P(A∪B) = P(K)+P(H)−P(K∩H). The King of Hearts is in BOTH groups — don't double count it!

A 17/52

B 4/13

C 16/52

D 1/4

Solution

P(King) = 4/52. P(Heart) = 13/52. P(King of Hearts) = 1/52.

P(King∪Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

17/52 = 4+13 without subtracting the overlap. The King of Hearts belongs to both groups and would be counted twice.

Probability · Misread Trap

A fair coin is tossed three times. What is the probability of getting exactly two heads?

💡 List outcomes with exactly 2 H's: HHT, HTH, THH — there are 3. Total outcomes = 2³ = 8.

A 1/4

B 3/8

C 1/2

D 1/8

Solution

Total outcomes = 2³ = 8.

Exactly 2 heads: HHT, HTH, THH → 3 outcomes.

P = 3/8.

1/4 = P(HH on first two) but then ignores the third toss. Always list the full sample space.

Probability · Word Problem

In a group of 50 people: 20 speak only English, 15 speak only French, 10 speak both, 5 speak neither.

A person is chosen at random. Given that they speak French, what is the probability they also speak English?

💡 n(French total) = "only French" + "both" = 15 + 10 = 25. n(French AND English) = 10 (both).

A 1/5

B 2/5

C 1/3

D 10/50

Solution

n(French) = 15 + 10 = 25. n(both) = 10.

P(English | French) = 10/25 = 2/5.

10/50 = P(both) overall — but we are told the person already speaks French, so restrict to the 25 French speakers.

Probability · Classic Hard Trap

Events A and B are such that P(A) = 0.5, P(B) = 0.4, P(A∪B) = 0.7.

Which statement is correct?

💡 First find P(A∩B) using the addition rule. Then check independence: is P(A)×P(B) = P(A∩B)?

A A and B are mutually exclusive

B A and B are independent

C A and B are neither independent nor mutually exclusive

D A and B are both independent and mutually exclusive

Solution

P(A∩B) = P(A)+P(B)−P(A∪B) = 0.5+0.4−0.7 = 0.2.

Mutually exclusive? P(A∩B) = 0? → 0.2 ≠ 0 → NOT mutually exclusive.

Independent? P(A)×P(B) = 0.5×0.4 = 0.2 = P(A∩B) ✓ → Independent!

A and B are independent but NOT mutually exclusive. These two properties are completely different concepts!

Probability &
Set Theory

Core Symbols

⚡ Memory Points

Solution

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Key Rules

⚡ Memory Points

Solution

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Addition Rules

⚡ Memory Points

Solution

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Conditional Probability

⚡ Memory Points

Solution

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🎉

Quiz Complete!

Probability &Set Theory

Core Symbols

⚡ Memory Points

Solution

Solution

Solution

Solution

Solution

Key Rules

⚡ Memory Points

Solution

Solution

Solution

Addition Rules

⚡ Memory Points

Solution

Solution

Solution

Conditional Probability

⚡ Memory Points

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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Solution

🎉

Quiz Complete!

Probability &
Set Theory