Mathematics Advanced • Year 12 • Module 5 • Lesson 1

Introduction to Probability

Practise HSC-style writing on sample spaces, Venn diagrams and the addition rule — including a 3-set extended response.

Master · Past-Paper Style

1. Short-answer questions

1.1 A standard 52-card deck is shuffled and one card is drawn. Let R = "card is red" and F = "card is a face card (J, Q, K)". Find P(R), P(F), P(R ∩ F) and P(R ∪ F), and state which probability rule you used at each step. 3 marks Band 3

1.2 In a class of 25 students, 14 study Music (M) and 11 study Drama (D). If 5 students study neither, find P(M ∩ D) and explain why your answer matches the Venn-diagram counts. 3 marks Band 3-4

1.3 A bag contains 4 red and 6 blue balls. Two balls are drawn one after the other without replacement. By listing or otherwise, find P(both same colour). 4 marks Band 4

Stuck on 1.3? Split the event into "both red" + "both blue" (mutually exclusive) and use multiplication along each branch.

2. Extended response

2.1 A survey of 200 Year 12 students records subject selection across three faculties:

120 study at least one Science subject (event S)
100 study at least one Mathematics subject (event M)
80 study a Language (event L)
60 study both Science and Mathematics
30 study Science and Language
25 study Mathematics and Language
15 study all three faculties

(a) Draw a 3-circle Venn diagram and fill in every region.
(b) Find the number of students who study none of these subjects, and hence find P(S ∪ M ∪ L).
(c) Verify your value of P(S ∪ M ∪ L) using the inclusion–exclusion formula for three events:
P(S ∪ M ∪ L) = P(S) + P(M) + P(L) − P(S ∩ M) − P(S ∩ L) − P(M ∩ L) + P(S ∩ M ∩ L).
(d) Hence find P(exactly one of S, M, L). 8 marks Band 5-6

Explicit marking criteria

Part (a) — 3 marks

• 1 mark — correct innermost region S ∩ M ∩ L = 15, and the three "exactly two" regions found by subtraction (S ∩ M only = 60 − 15 = 45; S ∩ L only = 30 − 15 = 15; M ∩ L only = 25 − 15 = 10).

• 1 mark — correct three "only" regions: S only = 120 − (45 + 15 + 15) = 45; M only = 100 − (45 + 10 + 15) = 30; L only = 80 − (15 + 10 + 15) = 40.

• 1 mark — neat, labelled diagram with all 7 internal regions and the "neither" tally.

Part (b) — 2 marks

• 1 mark — sums the 7 internal regions to 200 − 200 = 0 students with none. Wait — check: 45 + 30 + 40 + 45 + 15 + 10 + 15 = 200, so neither = 0.

• 1 mark — states P(S ∪ M ∪ L) = 200/200 = 1.

Part (c) — 2 marks

• 1 mark — substitutes correctly into the 3-set inclusion–exclusion formula.

• 1 mark — arrives at the same value (1.00) as part (b), with one line confirming consistency.

Part (d) — 1 mark — adds the three "only" regions: 45 + 30 + 40 = 115, giving P(exactly one) = 115/200 = 23/40.

Your response:

Always start a Venn-diagram problem from the innermost region (triple intersection) and work outwards by subtraction.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Red card / face card (3 marks)

Sample response. n(S) = 52. P(R) = 26/52 = 1/2 (equally-likely formula). P(F) = 12/52 = 3/13 (12 face cards). P(R ∩ F) = 6/52 = 3/26 (6 red face cards: J, Q, K of hearts and diamonds). By the addition rule, P(R ∪ F) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13.

Marking notes. 1 mark — correct P(R) and P(F). 1 mark — correct P(R ∩ F) with brief justification (e.g. "6 red face cards"). 1 mark — addition rule applied correctly with overlap subtracted. Common error: forgetting to subtract P(R ∩ F) and stating 38/52.

1.2 — Music and Drama (3 marks)

Sample response. Let x = n(M ∩ D). Venn regions: M only = 14 − x, D only = 11 − x, both = x, neither = 5. Total: (14 − x) + (11 − x) + x + 5 = 25 ⇒ 30 − x = 25 ⇒ x = 5. So P(M ∩ D) = 5/25 = 1/5. Venn check: M only = 9, D only = 6, both = 5, neither = 5 → 9 + 6 + 5 + 5 = 25 ✓.

Marking notes. 1 mark — sets up the "total = 25" equation correctly with all four regions. 1 mark — solves x = 5. 1 mark — converts to probability and verifies with Venn-region count. Common error: using P(M ∪ D) = P(M) + P(D) and forgetting the 5 "neither" students entirely.

1.3 — Two balls without replacement (4 marks)

Sample response. P(both red) = (4/10) × (3/9) = 12/90 = 2/15. P(both blue) = (6/10) × (5/9) = 30/90 = 1/3. "Both same colour" splits into two mutually exclusive events, so:
P(both same colour) = 2/15 + 1/3 = 2/15 + 5/15 = 7/15.

Marking notes. 1 mark — recognises sampling without replacement (denominator drops 10 → 9). 1 mark — P(both red) correct. 1 mark — P(both blue) correct. 1 mark — adds the two mutually exclusive events to give 7/15. Common error: using (4/10)(3/10) for "both red" — that's the with-replacement formula.

2.1 — Three-faculty survey (8 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) Venn diagram. Start from the innermost region and work outwards by subtraction. [Strategy note for marker.]

S ∩ M ∩ L = 15 (given) [1 mark — innermost region + correct "exactly two" regions from subtraction.]

S ∩ M only = 60 − 15 = 45 S ∩ L only = 30 − 15 = 15 M ∩ L only = 25 − 15 = 10

S only = 120 − (45 + 15 + 15) = 45 [1 mark — three "only" regions.]
M only = 100 − (45 + 10 + 15) = 30
L only = 80 − (15 + 10 + 15) = 40

Diagram: three overlapping circles labelled S, M, L with the seven internal counts entered (45, 30, 40, 45, 15, 10, 15) and the surrounding "neither" tally below. [1 mark — neat, fully labelled.]

(b) P(S ∪ M ∪ L). Sum of internal regions = 45 + 30 + 40 + 45 + 15 + 10 + 15 = 200. [1 mark — total of seven regions.]
So students studying at least one of S, M, L = 200, and "neither" = 200 − 200 = 0. Therefore P(S ∪ M ∪ L) = 200/200 = 1. [1 mark — probability stated.]

(c) Verification with inclusion–exclusion.

P(S ∪ M ∪ L) = P(S) + P(M) + P(L) − P(S∩M) − P(S∩L) − P(M∩L) + P(S∩M∩L)

= 120/200 + 100/200 + 80/200 − 60/200 − 30/200 − 25/200 + 15/200

= (120 + 100 + 80 − 60 − 30 − 25 + 15) / 200 = 200/200 = 1 ✓ [1 mark — correct substitution; 1 mark — matches part (b).]

(d) P(exactly one of S, M, L). Sum the three "only" regions:

P(exactly one) = (45 + 30 + 40)/200 = 115/200 = 23/40. [1 mark — uses the three "only" totals; not the marginal P(S) etc.]

Total: 8/8.

Band descriptors for marker.

Band 3: Sets up the Venn diagram but does not subtract the triple intersection correctly; finds P(S), P(M), P(L) only. ≈ 3-4 marks.

Band 4: Completes Venn diagram, finds P(S ∪ M ∪ L) correctly, but does not verify with inclusion–exclusion or compute "exactly one". ≈ 5-6 marks.

Band 5: All four parts attempted; minor arithmetic slip in inclusion–exclusion. ≈ 6-7 marks.

Band 6: All parts complete, with explicit verification step showing the two methods agree, and final part (d) distinguishing "exactly one" from the marginal totals. 8/8.