Topic 01
Set Theory
Notation, operations, Venn diagrams, and the inclusion-exclusion principle.
Quick Memory Point
UNION = OR (more) | INTERSECTION = AND (overlap) | COMPLEMENT = NOT (outside)
|A ∪ B| = |A| + |B| − |A ∩ B| ← always subtract the overlap once!
|A ∪ B| = |A| + |B| − |A ∩ B| ← always subtract the overlap once!
01
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5}.
What is A ∩ B?
What is A ∩ B?
A ∩ B = elements that belong to both A and B
✓ Correct Answer: B — {2, 4}
Intersection (∩) means elements in both sets. A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5}. The only numbers appearing in both are 2 and 4.
Intersection (∩) means elements in both sets. A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5}. The only numbers appearing in both are 2 and 4.
A ∩ B = {x : x ∈ A AND x ∈ B} = {2, 4}
Trap: choice A is just set A, choice C is A ∪ B, and D is B \ A (complement of A in B).
02
In a class of 30 students, 18 play soccer, 15 play basketball, and 8 play both.
How many students play neither sport?
How many students play neither sport?
|A ∪ B| = |A| + |B| − |A ∩ B|
Neither = Total − |A ∪ B|
Neither = Total − |A ∪ B|
✓ Correct Answer: C — 5
Step 1: |A ∪ B| = 18 + 15 − 8 = 25
Step 2: Neither = 30 − 25 = 5
Step 1: |A ∪ B| = 18 + 15 − 8 = 25
Step 2: Neither = 30 − 25 = 5
Neither = Total − (|Soccer| + |Basketball| − |Both|) = 30 − 25 = 5
Common mistake: forgetting to subtract the overlap, giving 30 − 33 (negative!) or just doing 30 − 18 − 15 + 8 = 5 (same result — good shortcut!).
03
U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 3, 5, 7}
Find A′ (the complement of A).
Find A′ (the complement of A).
✓ Correct Answer: B — {2, 4, 6, 8}
A′ = "everything in U but NOT in A." Since A contains the odd numbers, A′ contains the even numbers.
A′ = "everything in U but NOT in A." Since A contains the odd numbers, A′ contains the even numbers.
A′ = U \ A = {1,2,3,4,5,6,7,8} \ {1,3,5,7} = {2,4,6,8}
04
In the Venn diagram below, the numbers represent the count of elements in each region.
What is |A ∪ B|?
What is |A ∪ B|?
✓ Correct Answer: B — 28
A ∪ B includes all three regions inside the circles: only-A (12) + overlap (7) + only-B (9).
A ∪ B includes all three regions inside the circles: only-A (12) + overlap (7) + only-B (9).
|A ∪ B| = 12 + 7 + 9 = 28
The "5" outside both circles is in U but NOT in A ∪ B. Total |U| = 12 + 7 + 9 + 5 = 33 ← classic trap answer!
05
Which of the following is equal to (A ∪ B)′?
This is De Morgan's First Law — a very common exam question.
This is De Morgan's First Law — a very common exam question.
✓ Correct Answer: B — A′ ∩ B′
De Morgan's Laws (must memorize!):
Think of it this way: "NOT (A or B)" = "NOT A and NOT B."
De Morgan's Laws (must memorize!):
(A ∪ B)′ = A′ ∩ B′
(A ∩ B)′ = A′ ∪ B′
Memory trick: "UNION becomes INTERSECTION when you flip (complement) both sides."(A ∩ B)′ = A′ ∪ B′
Think of it this way: "NOT (A or B)" = "NOT A and NOT B."
Topic 02
Permutations & Combinations
Counting methods, factorial, nPr, nCr — and the tricky "when does order matter?" question.
Quick Memory Point
ORDER MATTERS → PERMUTATION nPr = n! / (n−r)!
ORDER DOESN'T MATTER → COMBINATION nCr = n! / (r! · (n−r)!)
Ask yourself: "Does swapping two items make a different outcome?" If YES → Permutation.
ORDER DOESN'T MATTER → COMBINATION nCr = n! / (r! · (n−r)!)
Ask yourself: "Does swapping two items make a different outcome?" If YES → Permutation.
06
How many ways can 5 students be arranged in a line?
n! = n × (n−1) × (n−2) × … × 1
✓ Correct Answer: C — 120
Arranging ALL n items in a line = n!
Arranging ALL n items in a line = n!
5! = 5 × 4 × 3 × 2 × 1 = 120
Order matters here (ABCDE ≠ BACDE), so we use permutation. All 5 positions are filled, so it's simply 5 factorial.
07
A team of 3 is chosen from 8 students. How many different teams are possible?
C(n, r) = n! / (r! × (n−r)!)
✓ Correct Answer: C — 56
A "team" has no fixed roles → order doesn't matter → Combination.
A "team" has no fixed roles → order doesn't matter → Combination.
C(8, 3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56
Trap: 336 is P(8,3) — the permutation answer (when order matters, e.g., captain/vice/secretary).
08
4 books are arranged on a shelf. 2 specific books (Book A and Book B) must always be next to each other. How many arrangements are possible?
✓ Correct Answer: D — 12
Glue trick: Treat A+B as one unit → now 3 units to arrange.
Glue trick: Treat A+B as one unit → now 3 units to arrange.
Step 1: Arrange 3 units: 3! = 6 ways
Step 2: A and B can swap within the pair: 2! = 2 ways
Total = 6 × 2 = 12
This "gluing together" trick works whenever items must be adjacent.
Step 2: A and B can swap within the pair: 2! = 2 ways
Total = 6 × 2 = 12
09
What is the value of C(10, 7)?
Hint: Use the symmetry property of combinations!
Hint: Use the symmetry property of combinations!
C(n, r) = C(n, n−r) ← Symmetry Property
✓ Correct Answer: D — 120
Use symmetry: C(10, 7) = C(10, 10−7) = C(10, 3)
Use symmetry: C(10, 7) = C(10, 10−7) = C(10, 3)
C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120
This is much faster than computing C(10, 7) directly. Always flip to the smaller side when r > n/2!
10
A password has exactly 3 digits (0–9) followed by 2 capital letters (A–Z). Repetition is allowed. How many passwords are possible?
Fundamental Counting Principle: multiply the choices at each step
✓ Correct Answer: C — 676,000
Multiply choices at each position (repetition allowed):
Multiply choices at each position (repetition allowed):
10 × 10 × 10 × 26 × 26 = 1000 × 676 = 676,000
Trap B (67,600) forgets one digit. Trap D uses 36 for letters instead of 26.
Topic 03
Basic Probability
Sample space, theoretical probability, complementary events, and tree diagrams.
Quick Memory Point
P(event) = favourable outcomes / total outcomes
P(A) + P(A′) = 1 → so P(NOT A) = 1 − P(A)
Probability is always between 0 (impossible) and 1 (certain).
P(A) + P(A′) = 1 → so P(NOT A) = 1 − P(A)
Probability is always between 0 (impossible) and 1 (certain).
11
A bag contains 4 red, 3 blue, and 5 green marbles. One marble is drawn at random. What is the probability of drawing a blue marble?
✓ Correct Answer: B — 1/4
Total marbles = 4 + 3 + 5 = 12
Total marbles = 4 + 3 + 5 = 12
P(blue) = 3 / 12 = 1/4
Always compute the total first! Trap A (1/3) uses total = 9 (incorrect), trap C mixes up values.
12
The probability that it rains on a given day is 0.35. What is the probability that it does NOT rain?
P(A′) = 1 − P(A)
✓ Correct Answer: C — 0.65
P(no rain) = 1 − P(rain) = 1 − 0.35 = 0.65
The complement rule is one of the most used tools in probability. Always check: P(A) + P(A′) must equal 1.
13
Two fair dice are rolled. What is the probability that the sum is greater than 9?
Total outcomes = 6 × 6 = 36
Total outcomes = 6 × 6 = 36
✓ Correct Answer: B — 1/6
Sums greater than 9 means sum = 10, 11, or 12.
Sum 10: (4,6),(5,5),(6,4) → 3 outcomes
Sum 11: (5,6),(6,5) → 2 outcomes
Sum 12: (6,6) → 1 outcome
Total favourable = 3 + 2 + 1 = 6
Sums greater than 9 means sum = 10, 11, or 12.
Sum 10: (4,6),(5,5),(6,4) → 3 outcomes
Sum 11: (5,6),(6,5) → 2 outcomes
Sum 12: (6,6) → 1 outcome
Total favourable = 3 + 2 + 1 = 6
P(sum > 9) = 6 / 36 = 1/6
14
A coin is flipped 3 times. What is the probability of getting at least one Head?
P(at least one) = 1 − P(none)
✓ Correct Answer: C — 7/8
Instead of listing all cases with heads, use the complement:
Instead of listing all cases with heads, use the complement:
P(at least one H) = 1 − P(no heads at all)
P(TTT) = (1/2)³ = 1/8
∴ P(at least one H) = 1 − 1/8 = 7/8
Trap B (3/8) is the probability of exactly one head, which is a different question. "At least one" almost always means use the complement!
P(TTT) = (1/2)³ = 1/8
∴ P(at least one H) = 1 − 1/8 = 7/8
15
A box has 5 red and 3 white balls. Two balls are drawn without replacement. What is the probability that both are red?
✓ Correct Answer: B — 5/14
Without replacement means the second probability changes!
Without replacement means the second probability changes!
P(1st red) = 5/8
P(2nd red | 1st was red) = 4/7 (only 4 red left, 7 balls total)
P(both red) = 5/8 × 4/7 = 20/56 = 5/14
Trap A (25/64) uses 5/8 × 5/8 — this is WITH replacement. Always ask: "Does the first draw affect the second?" Without replacement → YES, it changes!
P(2nd red | 1st was red) = 4/7 (only 4 red left, 7 balls total)
P(both red) = 5/8 × 4/7 = 20/56 = 5/14
Topic 04
Conditional & Combined Probability
Independent events, mutually exclusive events, and conditional probability.
Quick Memory Point
INDEPENDENT: P(A ∩ B) = P(A) × P(B) | "one doesn't affect the other"
MUTUALLY EXCLUSIVE: P(A ∩ B) = 0 | "can't happen together"
CONDITIONAL: P(A|B) = P(A ∩ B) / P(B) | "given B already happened"
MUTUALLY EXCLUSIVE: P(A ∩ B) = 0 | "can't happen together"
CONDITIONAL: P(A|B) = P(A ∩ B) / P(B) | "given B already happened"
16
P(A) = 0.4 and P(B) = 0.3, and A and B are independent. What is P(A ∩ B)?
If independent: P(A ∩ B) = P(A) × P(B)
✓ Correct Answer: C — 0.12
P(A ∪ B) = 0.4 + 0.3 − 0.12 = 0.58 (trap B). Trap A confuses "independent" with P(A) + P(B).
P(A ∩ B) = P(A) × P(B) = 0.4 × 0.3 = 0.12
Trap A (0.70) is P(A ∪ B) = 0.4 + 0.3 − 0.12 = 0.58... no wait, that's B. Let's be precise:P(A ∪ B) = 0.4 + 0.3 − 0.12 = 0.58 (trap B). Trap A confuses "independent" with P(A) + P(B).
17
Events A and B are mutually exclusive. P(A) = 0.25 and P(B) = 0.45. Find P(A ∪ B).
Mutually exclusive: P(A ∪ B) = P(A) + P(B)
✓ Correct Answer: B — 0.70
Mutually exclusive means P(A ∩ B) = 0 (they can't both happen).
Mutually exclusive means P(A ∩ B) = 0 (they can't both happen).
P(A ∪ B) = P(A) + P(B) − 0 = 0.25 + 0.45 = 0.70
Common confusion: mutually exclusive ≠ independent. If A and B are mutually exclusive, knowing A happened tells you B definitely did NOT → they are actually dependent!
18
In a group of 100 students: 60 study Maths, 50 study Science, and 30 study both. A student is chosen at random. Given that the student studies Maths, what is the probability they also study Science?
P(Science | Maths) = P(Science ∩ Maths) / P(Maths)
✓ Correct Answer: C — 0.50
Convert counts to probabilities: P(both) = 30/100 = 0.30, P(Maths) = 60/100 = 0.60
Convert counts to probabilities: P(both) = 30/100 = 0.30, P(Maths) = 60/100 = 0.60
P(Science | Maths) = P(both) / P(Maths) = 0.30 / 0.60 = 0.50
Think of it this way: among the 60 Maths students, 30 also do Science → 30/60 = 1/2 = 0.50. Conditional probability zooms in on a smaller sample space!
19
A student says: "P(A) = 0.6, P(B) = 0.7, and A and B are mutually exclusive." Is this possible?
✓ Correct Answer: B
If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B) = 0.6 + 0.7 = 1.3.
If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B) = 0.6 + 0.7 = 1.3.
But probability CANNOT exceed 1 — this is impossible!
For mutually exclusive events, we need P(A) + P(B) ≤ 1. Mutually exclusive events can only coexist if the total stays ≤ 1. This is a favourite IB "true/false" style question.
20
From a group of 10 people, a committee of 4 is formed. In how many ways can the committee be formed if a specific person (Alex) must be included?
Hint: If Alex is already chosen, how many spots and people remain?
Hint: If Alex is already chosen, how many spots and people remain?
✓ Correct Answer: C — 84
Alex is fixed in the committee. Now choose 3 more people from the remaining 9.
Alex is fixed in the committee. Now choose 3 more people from the remaining 9.
C(9, 3) = 9! / (3! × 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 504 / 6 = 84
Trap A (210) = C(10, 4) — forgetting that Alex is already selected. Trap D (126) = C(9, 4) — choosing 4 more instead of 3. When a person MUST be included, subtract 1 from both the group size AND the selection size.