Mathematics • Year 8 • Unit 4 • Lesson 12
Two-Way Tables
Build fluency with reading and completing two-way tables. One worked example, one guided fill-in with blanks, then eight independent problems from cell reading to relative frequency calculations.
1. I do — fully worked example
Read every line. Each step has a short reason.
Problem. 120 students were surveyed about Sport and Music. The table below has one cell missing. Find it, then calculate two relative frequencies.
Music No Music Row Total
Sport 35 ? 80
No Sport 25 15 40
Col Total 60 60 120
Step 1 — Find the missing cell using the row total.
Sport row: 35 + ? = 80 ⇒ ? = 80 − 35 = 45
Reason: every row total equals the sum of the cells in that row. Solve like a puzzle.
Step 2 — Check using the column total.
No Music column: 45 + 15 = 60 ✓ (matches column total)
Reason: always cross-check with the other margin — both should agree.
Step 3 — Calculate two relative frequencies.
P(Sport AND Music) = 35 ÷ 120 ≈ 0.292 = 29.2%
P(Music | Sport) = 35 ÷ 80 = 0.4375 = 43.75% (conditional: row total in denominator)
Reason: relative frequency uses grand total; conditional probability uses the row (or column) total.
Answer: Missing cell = 45. P(Sport AND Music) ≈ 29.2%. P(Music given Sport) ≈ 43.75%.
2. We do — fill in the missing steps
Same shape as Section 1, but the working is faded. Fill in each blank. 5 marks
Problem. 200 people were surveyed about coffee and exercise habits. Complete the table and answer the questions.
Exercise No Exercise Row Total
Coffee 60 40 ?
No Coffee 30 ? 100
Col Total ? ? 200
Step 1 — Find each missing value.
Coffee row total: 60 + 40 = ______
No Coffee, No Exercise cell: 100 − 30 = ______
Exercise column total: 60 + 30 = ______
No Exercise column total: 40 + ______ = ______
Step 2 — Check that column totals add to grand total.
______ + ______ = 200 ✓
Step 3 — Calculate two probabilities:
P(Coffee AND Exercise) = 60 ÷ 200 = ______ = ______%
P(Exercise | Coffee) = 60 ÷ ______ = ______ ≈ ______%
3. You do — independent practice
Show your working in the space under each problem. Foundation → Standard → Extension.
Foundation — read cells and totals
Use this table for 3.1-3.4. 100 students surveyed on pet ownership and preferred animal type:
Cats Dogs Row Total
Has Pet 18 27 45
No Pet 20 35 55
Col Total 38 62 100
3.1 How many students have a pet AND prefer cats? 1 mark
3.2 How many students prefer dogs (regardless of pet ownership)? 1 mark
3.3 What is the grand total? 1 mark
3.4 Define joint frequency in one sentence. 1 mark
Standard — relative frequency and missing cells
3.5 Using the pets table above, calculate (a) P(Has Pet AND prefers Cats) as a percentage, and (b) P(prefers Dogs) as a percentage. 2 marks
3.6 Complete the table by finding the missing values labelled a, b, c, d.
Bus Walk Row Total
Junior a 24 64
Senior 22 b 46
Col Total c d 110 2 marks
Extension — conditional probability
3.7 Using the pets table, calculate P(prefers Cats | Has Pet) — the probability that someone prefers cats given they have a pet. Show the calculation including the correct denominator. 2 marks
3.8 Using the pets table, calculate P(prefers Cats | No Pet) and compare it to your answer for 3.7. Is "preferring cats" more common among pet owners or non-pet owners? Use the percentages to justify. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (coffee/exercise)
Coffee row total = 100. No Coffee + No Exercise = 100 − 30 = 70. Exercise column total = 60 + 30 = 90. No Exercise column total = 40 + 70 = 110. Check: 90 + 110 = 200 ✓.
P(Coffee AND Exercise) = 60 ÷ 200 = 0.30 = 30%.
P(Exercise | Coffee) = 60 ÷ 100 = 0.60 = 60%.
3.1 — Pet AND cats
18 students (read the Has Pet row × Cats column cell directly).
3.2 — Prefer dogs
Dogs column total = 62 students.
3.3 — Grand total
100 students (bottom-right cell).
3.4 — Joint frequency
A joint frequency is a single cell count — the number of people who fall into BOTH categories simultaneously (e.g. Has Pet AND prefers Cats = 18).
3.5 — Relative frequencies
(a) P(Has Pet AND Cats) = 18 ÷ 100 = 18%.
(b) P(Dogs) = 62 ÷ 100 = 62%.
3.6 — Complete the table
a = 64 − 24 = 40. b = 46 − 22 = 24. c = 40 + 22 = 62. d = 24 + 24 = 48. Check: 62 + 48 = 110 ✓ and 64 + 46 = 110 ✓.
3.7 — Conditional: cats given pet
P(Cats | Has Pet) = (cell where Has Pet meets Cats) ÷ (Has Pet row total) = 18 ÷ 45 = 0.40 = 40%.
3.8 — Conditional: cats given no pet
P(Cats | No Pet) = 20 ÷ 55 ≈ 0.364 = 36.4%. Compared to 3.7 (40%), preferring cats is slightly more common among pet owners (40% vs 36.4%). The difference is small — only about 3.6 percentage points — so the link between pet ownership and cat preference is weak in this data.