Mathematics • Year 8 • Unit 4 • Lesson 17

Venn Diagrams in the Real World

Apply Venn diagrams and the addition rule to real situations: streaming-service overlap, school clubs, gym membership data, news reading habits, and dietary surveys.

Apply · Real-World Maths

1. Word problems

Each problem uses Venn-diagram skills from Lesson 17 — start with the intersection, work outward, then read off probabilities. Show your working.

1.1 — Streaming subscriptions. A survey of 80 households asks: 50 have Netflix (N), 38 have Disney+ (D), and 22 have both.

(a) Fill in the four regions of the Venn diagram (N only, both, D only, neither).
(b) How many have at least one of the two services?
(c) Find P(neither service). 4 marks

Stuck? Start by putting 22 in the overlap, then subtract: N only = 50 − 22, D only = 38 − 22.

1.2 — School clubs. Of 50 Year 8 students, 24 are in the chess club (C), 18 are in the debating club (D), and 10 are in neither.

(a) How many are in at least one club?
(b) Use the addition rule to find how many are in BOTH clubs.
(c) Find P(chess only). 3 marks

Stuck? Let n(both) = x. Then n(C ∪ D) = n(C) + n(D) − n(C ∩ D) → 40 = 24 + 18 − x.

1.3 — Gym data. A gym has 200 members. 120 use weights (W), 90 use cardio machines (C), and 40 use both.

(a) Fill in the four regions of the Venn diagram.
(b) Find P(uses weights only).
(c) Verify the addition rule numerically: show that P(W) + P(C) − P(W ∩ C) = P(W ∪ C). 4 marks

Stuck on (c)? Calculate each fraction using 200 as the denominator and confirm both sides simplify to the same value.

1.4 — Reading habits. A survey of 100 adults asked: 45 read news online (O), 30 read a print newspaper (P), and 35 read neither.

(a) How many read at least one form?
(b) Find n(O ∩ P).
(c) Find P(reads print only). 3 marks

Stuck? "At least one" = total − neither. Then use the addition rule to find the overlap.

1.5 — Dietary survey. Of 60 Year 8 students, 22 are vegetarian (V), 14 are gluten-free (G), and 4 are both vegetarian AND gluten-free.

(a) Are V and G mutually exclusive? Justify.
(b) Find n(neither).
(c) A student is picked at random. Find P(V only OR G only). 3 marks

Stuck on (a)? Mutually exclusive means n(V ∩ G) = 0. What does the data tell us?

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate writes: "P(A ∪ B) = P(A) + P(B). I added them together so I added P(A) and P(B) directly." Their answer is wrong whenever the events overlap. In your own words explain (i) why simply adding P(A) and P(B) can give a value greater than 1 when the events overlap, (ii) why the addition rule subtracts P(A ∩ B), and (iii) describe what changes if the events are mutually exclusive. Use the term intersection somewhere in your answer.

Stuck? Revisit lesson § "The Addition Rule" — when we add P(A) + P(B) directly, the intersection is counted in BOTH P(A) and P(B), so it is counted twice.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Streaming subscriptions

(a) N only = 50 − 22 = 28; Both = 22; D only = 38 − 22 = 16; Neither = 80 − 28 − 22 − 16 = 14. Check: 28 + 22 + 16 + 14 = 80 ✓.
(b) At least one = 80 − 14 = 66.
(c) P(neither) = 14 / 80 = 7/40.

1.2 — School clubs

(a) At least one = 50 − 10 = 40.
(b) Using the addition rule: 40 = 24 + 18 − n(both), so n(both) = 42 − 40 = 2.
(c) Chess only = 24 − 2 = 22. P(chess only) = 22 / 50 = 11/25.

1.3 — Gym data

(a) W only = 120 − 40 = 80; Both = 40; C only = 90 − 40 = 50; Neither = 200 − 80 − 40 − 50 = 30. Check: 80 + 40 + 50 + 30 = 200 ✓.
(b) P(weights only) = 80 / 200 = 2/5.
(c) P(W) + P(C) − P(W ∩ C) = 120/200 + 90/200 − 40/200 = 170/200 = 17/20. Direct count: P(W ∪ C) = (80 + 40 + 50)/200 = 170/200 = 17/20 ✓ — both sides equal.

1.4 — Reading habits

(a) At least one = 100 − 35 = 65.
(b) 65 = 45 + 30 − n(O ∩ P), so n(O ∩ P) = 75 − 65 = 10.
(c) Print only = 30 − 10 = 20. P(print only) = 20 / 100 = 1/5.

1.5 — Dietary survey

(a) Not mutually exclusive — n(V ∩ G) = 4, which is not zero, so the events overlap.
(b) V only = 22 − 4 = 18; G only = 14 − 4 = 10; Neither = 60 − 18 − 4 − 10 = 28.
(c) "V only OR G only" = 18 + 10 = 28 students. P = 28 / 60 = 7/15.

2.1 — Explain your thinking (sample response)

Simply adding P(A) and P(B) can give a value greater than 1 because every member of the intersection is counted twice — once as part of A and once as part of B. For example, if P(A) = 0.6 and P(B) = 0.5, adding gives 1.1 which is impossible. The addition rule corrects this by subtracting P(A ∩ B) once: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events the intersection is empty (P(A ∩ B) = 0), so no double-counting happens and the rule simplifies to P(A ∪ B) = P(A) + P(B). The subtraction term is what makes the rule work for overlapping events.

Marking: 1 mark for explaining the "greater than 1" problem; 1 mark for explaining double-counting using "intersection"; 1 mark for stating the addition rule; 1 mark for explaining the mutually-exclusive simplification.