IB Math · Probability & Combinatorics

Question 1 Easy

In a class of 30 students, 18 study French and 14 study Spanish. If 6 students study both, how many study at least one of the two languages?

💡KEY: "At least one" = Union. Use: n(F∪S) = n(F) + n(S) − n(F∩S)

A 26

B 32

C 20

D 36

📖 Explanation

Using the inclusion-exclusion principle:
n(F∪S) = n(F) + n(S) − n(F∩S) = 18 + 14 − 6 = 26

Common mistake: adding 18 + 14 = 32 without subtracting the overlap. Students who study both are counted twice — once in each group — so you must subtract them once.

Question 2 Medium

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

Find (A ∪ B)′.

💡TRICK: First find A∪B, then take the complement from U. Don't forget — the prime symbol (′) means "everything NOT in that set."

A {3, 5}

B {2, 8}

C {1, 2, 6, 7, 8}

D {1, 7}

📖 Explanation

Step 1: A∪B = {1, 3, 4, 5, 6, 7} (all elements in A or B)
Step 2: (A∪B)′ = U − (A∪B) = {1,2,3,4,5,6,7,8} − {1,3,4,5,6,7} = {2, 8}

Common mistake: confusing (A∪B)′ with A′∩B′. By De Morgan's Law, they are actually equal — but students often compute (A∩B)′ instead!

Question 3 Tricky

In a survey of 50 people: 30 like apples, 25 like bananas, 10 like neither.
How many people like both apples and bananas?

💡STEP 1: Find n(A∪B) first. People who like at least one = total − neither = 50 − 10 = 40.

A 5

B 15

C 10

D 20

📖 Explanation

n(A∪B) = 50 − 10 = 40 (those who like at least one fruit)
Apply inclusion-exclusion: n(A∩B) = n(A) + n(B) − n(A∪B) = 30 + 25 − 40 = 15

Key insight: always subtract the "neither" group first to find the union, then work backwards for the intersection.

Question 4 Easy

A restaurant menu has 4 starters, 6 main courses, and 3 desserts. How many different 3-course meals can be chosen?

A 13

B 72

C 36

D 48

📖 Explanation

AND (choosing a starter AND a main AND a dessert) → multiply:
4 × 6 × 3 = 72

Remember: when you make independent choices, multiply. Only add when you're choosing one or the other (mutually exclusive options).

Question 5 Medium

How many 3-digit numbers can be formed from the digits {1, 2, 3, 4, 5} if repetition is NOT allowed?

💡VISUALIZE: __ × __ × __ → 1st digit: 5 choices, 2nd: 4 (one used), 3rd: 3 (two used)

A 125

B 15

C 60

D 30

📖 Explanation

No repetition → each digit used at most once.
1st position: 5 choices
2nd position: 4 remaining choices
3rd position: 3 remaining choices
Total = 5 × 4 × 3 = 60

If repetition were allowed: 5 × 5 × 5 = 125. Don't mix these up!

Question 6 Tricky

How many ways can 4 boys and 3 girls sit in a row if all girls must sit together?

💡BLOCK METHOD: Treat all girls as ONE block → arrange (4 boys + 1 block) = 5 units, then arrange girls inside the block.

A 720

B 5040

C 144

D 864

📖 Explanation

Step 1: Treat the 3 girls as 1 block → now arrange 5 units (4 boys + 1 girl-block) = 5! = 120 ways
Step 2: Arrange the 3 girls within their block = 3! = 6 ways
Total = 120 × 6 = 720

Classic trap: many students forget to multiply by the internal arrangements of the block. Both steps are essential!

Question 7 Easy

In how many ways can 5 different books be arranged on a shelf?

A 25

B 10

C 120

D 60

📖 Explanation

Arranging all 5 different items in order = 5! = 5 × 4 × 3 × 2 × 1 = 120

This is a permutation of all n items. The "arrangement" keyword signals that order matters — use factorial.

Question 8 Medium

How many ways can 3 people be selected from a group of 8 and arranged in a line?
(Order matters — 1st, 2nd, 3rd place)

₈P₃ = 8! / (8−3)! = 8! / 5!

A 56

B 336

C 512

D 24

📖 Explanation

₈P₃ = 8 × 7 × 6 = 336

Why 8×7×6? 1st place: 8 choices, 2nd place: 7 remaining, 3rd place: 6 remaining.
Note: If order didn't matter, we'd divide by 3! = 6, giving ₈C₃ = 56.

Question 9 Tricky

How many distinct arrangements are there of the letters in the word MISSISSIPPI?

💡IDENTICAL ITEMS: Total letters = 11. Count repeats: M×1, I×4, S×4, P×2. Divide by each group's factorial.

A 39,916,800

B 34,650

C 831,600

D 1,663,200

📖 Explanation

Letters: M=1, I=4, S=4, P=2 → Total = 11
Distinct arrangements = 11! / (1! × 4! × 4! × 2!)
= 39,916,800 / (1 × 24 × 24 × 2)
= 39,916,800 / 1,152 = 34,650

Key: repeated letters create identical arrangements, so divide them out.

Question 10 Medium

In how many ways can 6 people be seated around a circular table?

💡CIRCULAR: Fix one person to remove rotational duplicates → arrange remaining (n−1) people.

A 720

B 120

C 360

D 5040

📖 Explanation

Circular arrangements = (n−1)! = (6−1)! = 5! = 120

Why? In a circle, rotating everyone one seat gives the same arrangement. By fixing one person's seat, we eliminate 6 identical rotations. So 6!/6 = 5! = 120.
Common mistake: answering 6! = 720 (forgetting the circular adjustment).

Question 11 Easy

In how many ways can a committee of 3 people be chosen from 7 candidates?

₇C₃ = 7! / (3! × 4!)

A 210

B 35

C 21

D 42

📖 Explanation

₇C₃ = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35

"Committee" means order doesn't matter — choosing {Alice, Bob, Carol} is the same as {Carol, Alice, Bob}. That's why we divide by r! to remove ordered duplicates.

Question 12 Medium

A student must choose 4 questions from a paper with 6 questions in Part A and 4 questions in Part B.
They must choose at least 2 from Part A. How many ways can they do this?

💡CASES: Split into: exactly 2 from A, exactly 3 from A, exactly 4 from A. Add all cases.

A 186

B 210

C 156

D 120

📖 Explanation

Case 1 — 2 from A, 2 from B: ₆C₂ × ₄C₂ = 15 × 6 = 90
Case 2 — 3 from A, 1 from B: ₆C₃ × ₄C₁ = 20 × 4 = 80
Case 3 — 4 from A, 0 from B: ₆C₄ × ₄C₀ = 15 × 1 = 15
Total = 90 + 80 + 15 = 185

Note: This is a classic IB-style case-split problem. Always check "at least" means ≥, so include all valid case counts.

Question 13 Tricky

Which of the following is equal to ⁿCᵣ + ⁿCᵣ₋₁?
(This is Pascal's Identity)

A ⁿ⁺¹Cᵣ

B ⁿCᵣ₊₁

C 2·ⁿCᵣ

D ⁿ⁺¹Cᵣ₋₁

📖 Explanation

Pascal's Identity: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

This is the foundation of Pascal's Triangle — each entry is the sum of the two entries above it.
Example: ₅C₂ + ₅C₁ = 10 + 5 = 15 = ₆C₂ ✓

Question 14 Easy

A bag contains 5 red, 3 blue, and 2 green marbles. What is the probability of picking a ball that is NOT red?

A 1/2

B 5/10

C 3/10

D 1/5

📖 Explanation

Total marbles = 5 + 3 + 2 = 10
P(red) = 5/10 = 1/2
P(NOT red) = 1 − 1/2 = 1/2

Or directly: not-red = 3 blue + 2 green = 5, so P(NOT red) = 5/10 = 1/2.
Note: Both A and B equal 1/2, but 1/2 is the simplified form — always simplify.

Question 15 Medium

Two fair dice are rolled. What is the probability that the sum equals 7?

💡LIST SYSTEMATICALLY: Total outcomes = 6×6 = 36. Count pairs that sum to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).

A 1/6

B 7/36

C 1/12

D 5/36

📖 Explanation

Favourable pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs
Total outcomes: 6 × 6 = 36
P(sum = 7) = 6/36 = 1/6

Fun fact: 7 is the most likely sum with two dice. 6 and 8 each have 5 ways, 5 and 9 have 4 ways.

Question 16 Medium

Events A and B are independent. If P(A) = 0.4 and P(B) = 0.3,
find P(A ∪ B).

💡INDEPENDENT ≠ MUTUALLY EXCLUSIVE! Independent means P(A∩B) = P(A)×P(B). Mutually exclusive means P(A∩B) = 0.

A 0.70

B 0.12

C 0.58

D 0.82

📖 Explanation

Since independent: P(A∩B) = P(A) × P(B) = 0.4 × 0.3 = 0.12
P(A∪B) = P(A) + P(B) − P(A∩B) = 0.4 + 0.3 − 0.12 = 0.58

Trap answer: 0.70 = 0.4 + 0.3, which forgets to subtract the overlap (only valid if mutually exclusive).

Question 17 Tricky

A box has 4 red and 6 blue balls. Two balls are drawn without replacement.
Find the probability that both are red.

💡WITHOUT REPLACEMENT: The denominator changes on the 2nd draw! 10 total → 9 remaining after 1st draw.

A 16/100

B 2/15

C 4/25

D 1/6

📖 Explanation

P(1st red) = 4/10
P(2nd red | 1st was red) = 3/9 (only 3 red left, 9 total left)
P(both red) = (4/10) × (3/9) = 12/90 = 2/15

Common mistake: using 4/10 × 4/10 = 16/100 — this assumes replacement (independent draws). Without replacement, the pool shrinks!

Question 18 Tricky

In a class, 60% of students passed Math, 50% passed Science, and 30% passed both.
A student is selected at random. Given they passed Math, what is the probability they also passed Science?

💡CONDITIONAL PROBABILITY: P(S|M) = P(S∩M) / P(M). The "|" symbol means "given that."

A 0.50

B 0.30

C 0.18

D 0.60

📖 Explanation

P(S|M) = P(S∩M) / P(M) = 0.30 / 0.60 = 0.50

Interpretation: Of all students who passed Math (60%), half of them (30%) also passed Science. So the conditional probability is 0.3/0.6 = 1/2.
Common mistake: answering P(S) = 0.50 directly without using the conditional formula.

Question 19 Medium

A fair coin is flipped 3 times. What is the probability of getting exactly 2 heads?

💡BINOMIAL APPROACH: How many ways to choose which 2 of 3 flips are heads? Use ₃C₂. Then multiply by the probability of each outcome.

A 1/4

B 3/8

C 1/8

D 1/2

📖 Explanation

Favourable outcomes (HHT, HTH, THH) = ₃C₂ = 3 ways
Total outcomes = 2³ = 8
P(exactly 2 heads) = 3/8

Or: P = ₃C₂ × (1/2)² × (1/2)¹ = 3 × 1/4 × 1/2 = 3/8

Question 20 Tricky

A class has 12 boys and 8 girls. A committee of 5 is chosen at random.
What is the probability that the committee contains exactly 3 boys and 2 girls?

💡HYPERGEOMETRIC SETUP: P = (ways to pick 3 boys × ways to pick 2 girls) ÷ (total ways to pick any 5)

A 1540/2584

B 220/1001

C 6160/15504

D 3/10

📖 Explanation

Ways to pick 3 boys from 12: ₁₂C₃ = 220
Ways to pick 2 girls from 8: ₈C₂ = 28
Favourable = 220 × 28 = 6,160
Total ways to pick 5 from 20: ₂₀C₅ = 15,504
P = 6,160 / 15,504 = 6160/15504 = 385/969 ≈ 0.397

Simplify: 6160/15504 = 385/969. Answer C is the correct unsimplified form.