IB Math · Probability & Combinatorics

1

Set Theory

In a class of 30 students, 18 play football, 12 play basketball, and 5 play both. How many students play at least one sport?

UNION FORMULA: add both, subtract overlap

Worked example: If 10 like cats, 8 like dogs, 3 like both → n(C∪D) = 10 + 8 − 3 = 15

💡 Explanation

1

Use the Union Formula: |A∪B| = |A| + |B| − |A∩B|

2

= 18 + 12 − 5 = 25

3

5 students play both, so we subtract them once to avoid double-counting.

2

Set Theory

If U = {1,2,3,4,5,6,7,8}, A = {2,4,6,8}, and B = {1,3,5,7}, what is A ∩ B?

INTERSECTION = only what's in BOTH

💡 Explanation

1

A contains even numbers, B contains odd numbers.

2

No number is both even AND odd, so A ∩ B = ∅.

3

When two sets share no elements, they are called disjoint sets.

3

Set Theory

A = {a, b, c, d}. How many subsets does A have (including ∅ and A itself)?

SUBSETS of n elements = 2ⁿ

Quick tip: Every element is either IN or OUT of a subset → 2 choices per element → 2n total subsets.

💡 Explanation

1

Formula: number of subsets = 2ⁿ where n = number of elements.

2

|A| = 4, so 2⁴ = 16 subsets.

3

These include ∅, {a}, {b}, {c}, {d}, {a,b}, … up to {a,b,c,d}.

4

Probability

A bag has 4 red, 3 blue, and 5 green marbles. One marble is drawn at random. What is P(not green)?

COMPLEMENT: P(A') = 1 − P(A)

💡 Explanation

1

Total marbles = 4+3+5 = 12

2

P(green) = 5/12

3

P(not green) = 1 − 5/12 = 7/12

5

Probability

A fair die is rolled. What is P(even OR greater than 4)?

P(A∪B) = P(A)+P(B)−P(A∩B) → avoid DOUBLE-COUNT

Trap alert: The number 6 is BOTH even AND greater than 4. Students often forget to subtract the overlap!

💡 Explanation

1

Even numbers: {2, 4, 6} → P(even) = 3/6

2

Greater than 4: {5, 6} → P(>4) = 2/6

3

Overlap (even AND >4): {6} → P = 1/6

4

P(A∪B) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3

6

Probability

Two fair coins are flipped. What is P(at least one Head)?

AT LEAST ONE = 1 − P(NONE)

💡 Explanation

1

Sample space: {HH, HT, TH, TT} → 4 outcomes

2

P(no heads) = P(TT) = 1/4

3

P(at least one head) = 1 − 1/4 = 3/4

7

Conditional Prob.

In a group, P(A) = 0.6, P(B) = 0.5, P(A∩B) = 0.3. What is P(A|B)?

P(A|B) = P(A∩B) / P(B) → "given B happened"

Read "A|B" as: "Probability of A given that B has already occurred." You zoom into the B universe.

💡 Explanation

1

Formula: P(A|B) = P(A∩B) / P(B)

2

= 0.3 / 0.5 = 0.6

3

Interesting: P(A|B) = P(A) = 0.6, which means A and B are independent!

8

Permutation

In how many ways can 5 students line up for a photo?

ALL objects in a LINE = n! (n factorial)

Example: 3 students → 3! = 3×2×1 = 6 ways

💡 Explanation

1

5! = 5 × 4 × 3 × 2 × 1

2

= 20 × 6 = 120

3

Each position has one fewer choice: 5 choices for 1st, 4 for 2nd, 3 for 3rd…

9

Permutation

How many 3-letter arrangements can be made from the letters in "MATH" (no repetition)?

⁴P₃ = 4! / (4−3)! = choosing 3 from 4 in ORDER

💡 Explanation

1

⁴P₃ = 4! / (4−3)! = 4! / 1! = 24/1 = 24

2

Or think: 4 choices × 3 choices × 2 choices = 4×3×2 = 24

10

Permutation

How many different 4-digit PIN codes can be created if repetition is allowed?

REPETITION ALLOWED: n choices each time → nʳ

Trap: When repetition is allowed, you do NOT divide by anything. Each digit independently has 10 choices (0–9).

💡 Explanation

1

Digits: 0–9 → 10 options per digit

2

4 digits with repetition: 10⁴ = 10,000

3

PINs range from 0000 to 9999 → exactly 10,000 codes.

11

Permutation ★

6 people sit in a circle. How many distinct arrangements are there?

CIRCULAR PERMUTATION = (n−1)! → fix one person, arrange the rest

Why n−1? In a circle, rotating everyone one seat looks the same. So we fix one person and arrange the remaining 5.

💡 Explanation

1

Circular permutation formula: (n−1)!

2

(6−1)! = 5! = 5×4×3×2×1 = 120

3

Common mistake: using 6! = 720 (that's for a LINE, not a circle).

12

Combination

A team of 3 is chosen from 8 candidates. How many different teams are possible?

TEAM = no order → ⁸C₃ = 8!/(3!·5!)

💡 Explanation

1

⁸C₃ = 8! / (3! × 5!) = (8×7×6) / (3×2×1)

2

= 336 / 6 = 56

3

Note: ⁸P₃ = 336, but since order doesn't matter, divide by 3! = 6.

13

Combination ★

A committee of 4 is formed from 5 men and 4 women. How many committees include exactly 2 women?

RESTRICTED COMBO: split into groups, multiply

Strategy: Choose 2 from 4 women AND 2 from 5 men separately, then multiply. Multiplication = "AND".

💡 Explanation

1

Choose 2 women from 4: ⁴C₂ = 4!/(2!·2!) = 6

2

Choose 2 men from 5: ⁵C₂ = 5!/(2!·3!) = 10

3

Total = 6 × 10 = 60 committees

14

Combination

What is the value of ⁷C₀?

nC0 = 1 ALWAYS — there's only ONE way to choose nothing

💡 Explanation

1

⁷C₀ = 7! / (0! × 7!) = 1/1 = 1

2

Recall: 0! = 1 by definition.

3

Intuitively: there is exactly one way to choose nothing — the empty selection.

15

Perm vs Comb ★★

Which scenario uses a Combination (not a Permutation)?

ARRANGE → Permutation | SELECT → Combination

Memory trick: "PERM" sounds like "Performance" → order matters on stage. "COMBO" sounds like a meal deal → just getting items, order irrelevant.

💡 Explanation

1

A: "arranging" = order matters → Permutation ❌

2

B: different passwords, order matters (1234 ≠ 4321) → Permutation ❌

3

C: pizza toppings — {cheese, mushroom, olive} = same pizza regardless of pick order → Combination ✓

4

D: gold ≠ silver ≠ bronze, so order matters → Permutation ❌

16

Probability

Two cards are drawn without replacement from a standard 52-card deck. What is P(both are Aces)?

WITHOUT REPLACEMENT: denominators decrease each draw

Trap: After drawing 1 Ace, only 3 Aces remain out of 51 cards — NOT 4 out of 52!

💡 Explanation

1

P(1st Ace) = 4/52

2

P(2nd Ace | 1st was Ace) = 3/51 (one Ace and one card removed)

3

P(both Aces) = (4/52) × (3/51) = 12/2652 = 1/221

17

Independence ★★

P(A) = 0.4, P(B) = 0.3, P(A∩B) = 0.12. Are A and B independent?

INDEPENDENT if P(A∩B) = P(A) × P(B)

Check: If P(A)·P(B) ≠ P(A∩B), the events are dependent — knowing one affects the other.

💡 Explanation

1

Test: P(A) × P(B) = 0.4 × 0.3 = 0.12

2

P(A∩B) = 0.12 ← given

3

Since they're equal, A and B are independent. ✓

4

Bonus: Independent events can still overlap! They are NOT the same as mutually exclusive.

18

Combination

A student must answer 4 questions out of 6 on an exam. One specific question is compulsory. How many ways can the student choose their questions?

FIX compulsory first, then CHOOSE from the rest

💡 Explanation

1

1 question is compulsory → already selected. Now choose 3 more from the remaining 5.

2

⁵C₃ = 5! / (3! × 2!) = (5×4) / (2×1) = 10

19

Probability ★★

A box has 3 red and 5 blue balls. Two are drawn with replacement. What is P(one red, one blue)?

WITH REPLACEMENT: order can vary → multiply × 2 for both orders

Two orders exist: (Red then Blue) OR (Blue then Red). Add both!

💡 Explanation

1

Total = 8 balls. With replacement, denominator stays 8 each time.

2

P(Red then Blue) = (3/8) × (5/8) = 15/64

3

P(Blue then Red) = (5/8) × (3/8) = 15/64

4

P(one of each) = 15/64 + 15/64 = 30/64 = 15/32

20

Combinations ★★★

How many ways can 5 letters be chosen from {A, B, C, D, E, F, G} if the letter A must be included and B must be excluded?

INCLUDE/EXCLUDE: fix A in, remove B, then choose freely from leftovers

Step by step: A is fixed → need 4 more. B is banned → 5 letters left ({C,D,E,F,G}). Choose 4 from 5.

💡 Explanation

1

Start with 7 letters: A,B,C,D,E,F,G

2

A is always included → placed in the group. Need 4 more.

3

B is excluded → remove from pool. Remaining: C,D,E,F,G (5 letters).

4

Choose 4 from 5: ⁵C₄ = 5! / (4!·1!) = 5