Mathematics • Year 9 • Unit 4 • Lesson 15

Sets, Venn Diagrams & the Addition Rule — Mixed Challenge

Combine everything from Lessons 13-15 — sample spaces, complements, the addition rule and Venn diagrams. Choose the right tool, fix a classmate's mistake, and tackle an open-ended challenge.

Master · Mixed Challenge

1. Mixed problems — choose the right tool

Decide which rule or diagram to use before you start writing. Show working. 3 marks each

1.1 In a class of 35, 20 study French, 18 study German and 12 study both. Find how many study neither.

1.2 P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find P(A ∪ B), P(A only) and P(neither A nor B).

1.3 A card is drawn from a 52-card deck. Find P(red ∪ face card). Face cards = J, Q, K (12 total).

1.4 A 12-sector spinner is numbered 1-12. Find P(even ∪ multiple of 3) by listing the sets first, then applying the addition rule.

1.5 In a Venn diagram, the regions show: A only = 18, B only = 14, A ∩ B = 9, neither = 9. Find the total in the universal set, then P(A), P(B) and P(A ∩ B).

1.6 A Year 9 cohort of 60 students is surveyed about hobbies: 25 do gaming (G), 20 do sport (S), 15 do drawing (D); 8 do both G and S; 6 do both G and D; 5 do both S and D; and 2 do all three. Draw a three-circle Venn diagram and find how many do at least one hobby and how many do none.

Stuck on 1.6? Fill the triple overlap (2) first, then the pairwise overlaps (each minus the 2 in the triple), then the singles. Use the three-set inclusion-exclusion idea.

2. Find the mistake

Another student has tried to find P(A ∪ B) given P(A) = 0.5, P(B) = 0.4 and "A and B can both occur with probability 0.2". Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks

Student's working:

Line 1:   P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2.

Line 2:   A and B are mutually exclusive because they can both happen with probability 0.2.

Line 3:   For mutually exclusive events, P(A ∪ B) = P(A) + P(B).

Line 4:   So P(A ∪ B) = 0.5 + 0.4 = 0.9.

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong.

(c) Write out the corrected working in full, including the corrected final answer for P(A ∪ B).

Stuck? Revisit lesson § "Misconceptions" — "mutually exclusive" actually means the events cannot both occur (intersection = 0). Here the intersection is 0.2, so the events are NOT mutually exclusive.

3. Open-ended challenge — design a Venn-diagram class

This question has more than one valid answer. 4 marks

3.1 Design a Year 9 class of 40 students studying two electives, Art (A) and Music (M), with all four of these properties at once:

  • The number doing at least one elective is 30.
  • Twice as many students do Art as do Music.
  • Every Music student also does Art (i.e. M ⊆ A — Music is a subset of Art).
  • The number who do neither is 10.

For your class:
(i) Write the number in each region: A only, M only, A ∩ M, neither.
(ii) Verify all four conditions hold.
(iii) Sketch (in words or ASCII) what the Venn diagram looks like — what's special about it because M ⊆ A?

Stuck? If M ⊆ A then "M only" = 0. With 30 doing at least one and 0 in "M only", A only + (A ∩ M) = 30. Use "twice as many in A as M" to find the split.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — French / German / 35 students

French only = 20 − 12 = 8. German only = 18 − 12 = 6. Inside = 8 + 12 + 6 = 26. Neither = 35 − 26 = 9.

1.2 — Addition rule

P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8.
P(A only) = P(A) − P(A ∩ B) = 0.6 − 0.3 = 0.3.
P(neither) = 1 − P(A ∪ B) = 1 − 0.8 = 0.2.

1.3 — P(red ∪ face)

Red = 26, face = 12, red ∩ face = 6 (the red J, Q, K of hearts and diamonds = 2 × 3 = 6).
P(red ∪ face) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13.

1.4 — P(even ∪ multiple of 3) on 1-12 spinner

Even = {2, 4, 6, 8, 10, 12} (6 outcomes). Multiples of 3 = {3, 6, 9, 12} (4 outcomes). Both = {6, 12} (2 outcomes).
P(even ∪ mult 3) = 6/12 + 4/12 − 2/12 = 8/12 = 2/3.

1.5 — Reverse Venn

Total = 18 + 14 + 9 + 9 = 50.
P(A) = (18 + 9) / 50 = 27/50 = 0.54.
P(B) = (14 + 9) / 50 = 23/50 = 0.46.
P(A ∩ B) = 9/50 = 0.18.

1.6 — Three-circle Venn (60 students)

Triple G ∩ S ∩ D = 2.
G ∩ S only = 8 − 2 = 6. G ∩ D only = 6 − 2 = 4. S ∩ D only = 5 − 2 = 3.
G only = 25 − (6 + 4 + 2) = 13. S only = 20 − (6 + 3 + 2) = 9. D only = 15 − (4 + 3 + 2) = 6.
At least one hobby = 13 + 9 + 6 + 6 + 4 + 3 + 2 = 43. None = 60 − 43 = 17.

2 — Find the mistake

(a) The mistake is on Line 2.
(b) "Mutually exclusive" means the two events cannot both occur — so P(A ∩ B) would have to equal 0. Here P(A ∩ B) = 0.2, so the events are not mutually exclusive and the simplified formula P(A) + P(B) does not apply. The full addition rule must be used instead.
(c) Corrected working: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.4 − 0.2 = 0.7.

3 — Open-ended challenge (sample solution)

M ⊆ A means "M only" = 0. With 10 doing neither, exactly 30 students do at least one. So Art only + (A ∩ M) = 30.

"Twice as many in A as M" means |A| = 2 × |M|. Let |M| = m. Then |A| = 2m. Since M ⊆ A, A ∩ M = M, so A ∩ M = m, and A only = 2m − m = m.

Total in at least one = (A only) + (A ∩ M) + (M only) = m + m + 0 = 2m = 30 → m = 15.

So: Art only = 15; A ∩ M = 15; M only = 0; Neither = 10. Total = 15 + 15 + 0 + 10 = 40 ✓. |A| = 30, |M| = 15, and 30 = 2 × 15 ✓.

The Venn diagram for M ⊆ A is special — the Music circle sits entirely inside the Art circle, with no "M only" sliver outside Art.

Marking: 1 mark for spotting M ⊆ A forces M only = 0; 1 mark for setting up 2m = 30; 1 mark for the four regional counts; 1 mark for the verbal/sketch description of the nested-circle Venn diagram.