Mathematics Advanced • Year 12 • Module 5 • Lesson 1

Introduction to Probability

Build procedural fluency in sample spaces, Venn-diagram regions, the addition rule and complement counting.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each definition with the correct word or symbol:

The set of all possible outcomes of an experiment is the ____________ space, written ______.

For equally likely outcomes, P(A) = ______ / ______.

The complement of A is written ______, and satisfies P(A) + P(A′) = ______.

Q1.2 State the symbol that means each of the following set operations:

"A or B (at least one)" → ______    "A and B (both)" → ______    "not A" → ______

Write the addition rule in full: P(A ∪ B) = _________________________________.

Q1.3 Two events A and B are mutually exclusive. State the value of P(A ∩ B) and the simplified form of P(A ∪ B).   P(A ∩ B) = ______    P(A ∪ B) = _________________

Stuck? Revisit lesson § Sample Space & Events and § Venn Diagrams.

2. Worked example, P(B ∪ T) for a sport survey

Follow each step. Reasons are in italics on the right.

Problem. In a class of 30 students, 18 play basketball (B), 15 play tennis (T) and 8 play both. Find P(B), P(T), P(B ∩ T) and P(B ∪ T).

Step 1, Identify the sample space size.

n(S) = 30

Reason: every student is one equally likely outcome.

Step 2, Single-event probabilities.

P(B) = 18/30 = 3/5    P(T) = 15/30 = 1/2

Reason: P(A) = n(A)/n(S).

Step 3, Intersection.

P(B ∩ T) = 8/30 = 4/15

Reason: 8 students belong to both events; they sit in the overlap of the Venn diagram.

Step 4, Apply the addition rule.

P(B ∪ T) = P(B) + P(T) − P(B ∩ T)

= 18/30 + 15/30 − 8/30 = 25/30 = 5/6

Reason: adding 18 + 15 double-counts the 8 students in the overlap, so we subtract them once.

Conclusion. P(B ∪ T) = 5/6. The Venn regions are: B only = 10, T only = 7, both = 8, neither = 5 (total 30). ✓

3. Faded example, fill in the missing steps

In a group of 40 students, 22 study Biology (B), 18 study Chemistry (C), and 10 study both. Find P(B ∪ C). 4 marks

Step 1, Sample space. n(S) = ______

Step 2, Single events. P(B) = ______ / 40 = ______    P(C) = ______ / 40 = ______

Step 3, Intersection. P(B ∩ C) = ______ / 40 = ______

Step 4, Addition rule.

P(B ∪ C) = ______ + ______ − ______ = ______ / 40 = ______

Step 5, Venn check. B only = ______, C only = ______, both = ______, neither = ______    Total = 40 ✓

Stuck? Revisit lesson § Worked Example (sport survey).

4. Graduated practice, calculate the probability

Show the substitution and final answer as a simplified fraction or 3-decimal-place decimal. Assume fair / unbiased equipment unless stated.

Foundation, single-event probabilities (4 questions)

QQuestionWorking (one line)Probability
4.1 1Roll one fair die. P(rolling a 4).
4.2 1Draw one card from a standard 52-card deck. P(heart).
4.3 1Roll one die. P(even number).
4.4 1Flip two fair coins. P(at least one head). List the sample space first.

Standard, typical HSC difficulty (6 questions)

Show one line of working in each box.

4.5 Roll two fair dice. Find P(sum = 7). Hint: list the favourable ordered pairs.    2 marks

4.6 Given P(A) = 0.4, P(B) = 0.3 and P(A ∩ B) = 0.15, find P(A ∪ B).    2 marks

4.7 A bag holds 5 red, 3 blue and 2 green marbles. One marble is drawn. Find P(red ∪ blue).    2 marks

4.8 A die is rolled. Find P(rolling a number that is even or greater than 4). Use the addition rule and watch for overlap.    2 marks

4.9 A fair coin is flipped four times. Use the complement to find P(at least one tail).    2 marks

4.10 In a survey of 80 people: 45 own a smartphone, 50 own a laptop and 30 own both. Find P(owns smartphone or laptop).    2 marks

Extension, reasoning beyond procedure (2 questions)

4.11 Two events satisfy P(A) = 0.6 and P(B) = 0.5. Find the smallest and largest possible values of P(A ∪ B), and describe geometrically (in terms of a Venn diagram) what each extreme corresponds to.    3 marks

4.12 Three events A, B, C divide the sample space into 8 disjoint regions. Sketch a 3-circle Venn diagram and label all 8 regions in set notation (e.g. A ∩ B ∩ C′).    3 marks

Stuck on 4.11? P(A ∪ B) is smallest when one event is a subset of the other; largest when there is no overlap.

5. Self-check the easy 3

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

How did this worksheet feel?

What I'll revisit before next class:

Answers, Do not peek before attempting

Q1.1, Definitions

Sample space, written S.   P(A) = n(A) / n(S).   Complement A′; P(A) + P(A′) = 1.

Q1.2, Symbols and addition rule

"A or B" → ;   "A and B" → ;   "not A" → A′.   P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Q1.3, Mutually exclusive

P(A ∩ B) = 0;   P(A ∪ B) = P(A) + P(B) (no overlap to subtract).

Q3, Faded example (Biology / Chemistry)

Step 1: n(S) = 40.
Step 2: P(B) = 22/40 = 11/20;   P(C) = 18/40 = 9/20.
Step 3: P(B ∩ C) = 10/40 = 1/4.
Step 4: P(B ∪ C) = 22/40 + 18/40 − 10/40 = 30/40 = 3/4.
Step 5: B only = 12, C only = 8, both = 10, neither = 10. Total = 40 ✓.

Q4.1, P(rolling a 4)

n(A) = 1, n(S) = 6, so P = 1/6.

Q4.2, P(heart)

13 hearts in 52 cards: P = 13/52 = 1/4.

Q4.3, P(even)

Favourable outcomes {2, 4, 6}, so P = 3/6 = 1/2.

Q4.4, P(at least one head, two coins)

S = {HH, HT, TH, TT}, n(S) = 4. Favourable: {HH, HT, TH}, so P = 3/4. (Equivalently, 1 − P(TT) = 1 − 1/4 = 3/4.)

Q4.5, P(sum = 7) on two dice

Favourable ordered pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), 6 pairs. P = 6/36 = 1/6.

Q4.6, Addition rule

P(A ∪ B) = 0.4 + 0.3 − 0.15 = 0.55.

Q4.7, P(red ∪ blue) marbles

Red and blue are mutually exclusive (one marble cannot be two colours), so P = 5/10 + 3/10 = 8/10 = 4/5.

Q4.8, P(even or > 4) on a die

Even = {2, 4, 6} so P(even) = 3/6. >4 = {5, 6} so P(>4) = 2/6. Overlap = {6}, P(overlap) = 1/6.
P(even ∪ >4) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3.

Q4.9, P(at least one tail in 4 flips)

P(no tails) = P(HHHH) = (1/2)⁴ = 1/16. So P(at least one tail) = 1 − 1/16 = 15/16.

Q4.10, Smartphone or laptop

P = 45/80 + 50/80 − 30/80 = 65/80 = 13/16.

Q4.11, Bounds on P(A ∪ B)

Largest: if A and B are mutually exclusive (no overlap), P(A ∪ B) = 0.6 + 0.5 = 1.1 but probabilities cannot exceed 1, so the largest feasible value is 1, with overlap forced to be at least 0.1. Smallest: if one event is contained inside the other, P(A ∪ B) = max(P(A), P(B)) = 0.6; geometrically, the smaller circle (B) sits entirely inside the larger (A), so the union equals the larger circle.

Q4.12, Eight regions of a 3-circle Venn

(i) A ∩ B′ ∩ C′   (ii) A′ ∩ B ∩ C′   (iii) A′ ∩ B′ ∩ C   (iv) A ∩ B ∩ C′   (v) A ∩ B′ ∩ C   (vi) A′ ∩ B ∩ C   (vii) A ∩ B ∩ C   (viii) A′ ∩ B′ ∩ C′ (outside all three).