Mathematics Advanced • Year 12 • Module 5 • Lesson 4

Independence and Mutual Exclusivity

Build the algebraic test for independence and the test for mutual exclusivity, and learn to classify event pairs without confusing the two concepts.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each algebraic test:

Independence: P(A ∩ B) = ________________ (or equivalently P(A | B) = ________)

Mutual exclusivity: P(A ∩ B) = ________

Q1.2 Independence is about ____________; mutual exclusivity is about ____________.

(Hint: information vs possibility.)

Q1.3 Can two events with positive probability be both independent and mutually exclusive? Yes / No (circle) Justify in one line: ______________________________________________________

Stuck? Revisit lesson § Key distinction (independence vs mutual exclusivity).

2. Worked example — testing "even" and "> 4" on a die

Problem. A fair die is rolled. Let A = "even" and B = "> 4". Are A and B (i) mutually exclusive? (ii) independent?

Step 1 — List outcomes and find single-event probabilities.

A = {2, 4, 6}, P(A) = 3/6 = 1/2

B = {5, 6}, P(B) = 2/6 = 1/3

Step 2 — Find the intersection.

A ∩ B = {6}, P(A ∩ B) = 1/6

Step 3 — Test mutual exclusivity.

P(A ∩ B) = 1/6 ≠ 0 ⇒ NOT mutually exclusive (outcome 6 is in both).

Step 4 — Test independence.

P(A) × P(B) = (1/2) × (1/3) = 1/6 = P(A ∩ B). ✓ INDEPENDENT.

Reason: the product of the marginals matches the joint probability exactly.

Conclusion. A and B are not mutually exclusive but are independent — a surprising but valid combination.

3. Faded example — king and heart in a card draw

One card is drawn from a standard 52-card deck. Let C = "king" and D = "heart". Test whether C and D are mutually exclusive and whether they are independent. 4 marks

Step 1 — Single-event probabilities.

P(C) = ______ / 52 = ______ P(D) = ______ / 52 = ______

Step 2 — Intersection.

C ∩ D = { ________________ }, P(C ∩ D) = ______ / 52 = ______

Step 3 — Mutually exclusive?

P(C ∩ D) = ______, so C and D are ______________________ mutually exclusive.

Step 4 — Independent?

P(C) × P(D) = ______ × ______ = ______; this is ______________ P(C ∩ D), so C and D are / are not independent (circle one).

Conclusion. C and D are ____________________________________.

Stuck? Revisit lesson § Worked Example (even and >4 on a die).

4. Graduated practice — classify each event pair

For each pair (A, B), test for mutual exclusivity and independence. Show one line of working for each test.

Foundation — single-die experiments (4 questions)

Q	Event pair	Mut. excl.?	Independent?	Brief justification (one line)
4.1 1	A = "even", B = "5" on a die
4.2 1	A = "≥ 3", B = "≤ 4" on a die
4.3 1	A = "even", B = "prime" on a die
4.4 1	A = "1", B = "6" on a die

Standard — typical HSC difficulty (6 questions)

4.5 P(A) = 0.3, P(B) = 0.4, P(A ∩ B) = 0.12. Are A and B independent? 2 marks

4.6 P(A) = 0.5, P(B) = 0.5, P(A ∩ B) = 0. Are A and B (a) mutually exclusive? (b) independent? 2 marks

4.7 A card is drawn. C = "face card" (J, Q, K), D = "red". Test whether C and D are independent and whether they are mutually exclusive. 2 marks

4.8 A spinner has sectors red (30°), blue (60°), green (90°) and yellow (180°). Are "lands on red" and "lands on blue" mutually exclusive? Independent? 2 marks

4.9 Two fair coins are flipped. A = "first is head", B = "both are heads". Test for independence and mutual exclusivity. 2 marks

4.10 P(A) = 0.6, P(B) = 0.7, P(A ∪ B) = 0.88. Find P(A ∩ B) and test for independence. 2 marks

Extension — proofs (2 questions)

4.11 Prove that if A and B are independent, then A and B′ are also independent. (Hint: P(A ∩ B′) = P(A) − P(A ∩ B).) 3 marks

4.12 Suppose P(A) > 0 and P(B) > 0. Prove that if A and B are mutually exclusive, they cannot be independent. 3 marks

Stuck on 4.12? Combine "P(A ∩ B) = 0" (mutually exclusive) with "P(A) · P(B) > 0" — they cannot both equal P(A ∩ B).

5. Self-check the easy 3

Tick the first three once you've verified your method works.

For 4.1 I noticed B = {5} contains no even number, so P(A ∩ B) = 0 and A, B are mutually exclusive.

For 4.2 I noted A ∩ B = {3, 4} so they share outcomes, hence not mutually exclusive.

For 4.3 I computed P(A) × P(B) = (1/2)(1/2) = 1/4 and compared with P(A ∩ B) = 1/3 — they differ, so dependent.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Tests

Independence: P(A ∩ B) = P(A) × P(B) (or P(A | B) = P(A)). Mutual exclusivity: P(A ∩ B) = 0.

Q1.2 — Conceptual distinction

Independence is about information (does knowing A tell you about B?); mutual exclusivity is about possibility (can A and B both happen?).

Q1.3 — Both at once?

No. If P(A), P(B) > 0, mutual exclusivity gives P(A ∩ B) = 0 but independence would require P(A ∩ B) = P(A) × P(B) > 0. Contradiction. Mutually exclusive events are maximally dependent — if A happens, B definitely does not.

Q3 — Faded example (king and heart)

Step 1: P(C) = 4/52 = 1/13; P(D) = 13/52 = 1/4.
Step 2: C ∩ D = {king of hearts}, P(C ∩ D) = 1/52.
Step 3: P(C ∩ D) = 1/52 ≠ 0, so C and D are not mutually exclusive.
Step 4: P(C) × P(D) = (1/13)(1/4) = 1/52 = P(C ∩ D), so C and D are independent.
Conclusion: not mutually exclusive and independent.

Q4.1 — "even" vs "= 5"

A ∩ B = ∅ (5 is not even), so P(A ∩ B) = 0 → mutually exclusive. P(A)·P(B) = (1/2)(1/6) = 1/12 ≠ 0 → not independent.

Q4.2 — "≥ 3" vs "≤ 4"

A ∩ B = {3, 4}, P(A ∩ B) = 2/6 = 1/3 ≠ 0 → not mutually exclusive. P(A) = 4/6 = 2/3, P(B) = 4/6 = 2/3, P(A)·P(B) = 4/9 ≠ 1/3 → not independent.

Q4.3 — "even" vs "prime"

Primes on a die: {2, 3, 5}; A ∩ B = {2}, so P(A ∩ B) = 1/6 ≠ 0 → not mutually exclusive. P(A)·P(B) = (1/2)(1/2) = 1/4 ≠ 1/6 → not independent.

Q4.4 — "= 1" vs "= 6"

A ∩ B = ∅ → mutually exclusive. P(A)·P(B) = (1/6)(1/6) = 1/36 ≠ 0 → not independent.

Q4.5 — Test independence with given P's

P(A)·P(B) = 0.3 × 0.4 = 0.12 = P(A ∩ B). ✓ → A and B are independent.

Q4.6 — Mutually exclusive with equal P

(a) P(A ∩ B) = 0 → mutually exclusive. (b) P(A)·P(B) = 0.25 ≠ 0 → not independent. (This is the "maximally dependent" case from the lesson.)

Q4.7 — Face card and red card

P(C) = 12/52 = 3/13; P(D) = 26/52 = 1/2; P(C ∩ D) = 6/52 = 3/26 (6 red face cards). P(C)·P(D) = (3/13)(1/2) = 3/26 = P(C ∩ D) → independent. P(C ∩ D) ≠ 0 → not mutually exclusive.

Q4.8 — Spinner colours

The pointer lands in exactly one colour, so red and blue cannot both happen on one spin: mutually exclusive. P(red) = 30/360 = 1/12; P(blue) = 60/360 = 1/6; product = 1/72 ≠ 0 → not independent.

Q4.9 — "First H" vs "Both H"

P(A) = 1/2 (first is H); P(B) = 1/4 (HH); A ∩ B = HH, P(A ∩ B) = 1/4. P(A)·P(B) = 1/8 ≠ 1/4 → not independent (knowing both are heads guarantees first is head). P(A ∩ B) = 1/4 ≠ 0 → not mutually exclusive.

Q4.10 — Find P(A ∩ B) then test

From P(A ∪ B) = P(A) + P(B) − P(A ∩ B): 0.88 = 0.6 + 0.7 − P(A ∩ B), so P(A ∩ B) = 0.42. Test: P(A)·P(B) = 0.6 × 0.7 = 0.42 = P(A ∩ B) → independent.

Q4.11 — Independent ⇒ A and B′ independent

P(A ∩ B′) = P(A) − P(A ∩ B) (partition of A by B and B′)
= P(A) − P(A)·P(B) (given independence)
= P(A)(1 − P(B)) = P(A) × P(B′). ▮

Q4.12 — Mutually exclusive (P>0) ⇒ dependent

Mutually exclusive: P(A ∩ B) = 0. Independence requires P(A ∩ B) = P(A) × P(B). Since P(A), P(B) > 0, P(A)·P(B) > 0 ≠ 0 = P(A ∩ B). The two conditions contradict, so A and B cannot be independent — they are dependent. ▮