Mathematics • Year 10 • Unit 4 • Lesson 19
Conditional Probability — Mixed Challenge
Pull together Lesson 19: the formula P(A|B) = P(A and B)/P(B), reading conditionals from two-way tables and Venn diagrams, and using P(A|B) = P(A) as the independence test. Spot a Year 10 mistake and design your own conditional probability puzzle.
1. Mixed problems
Show working and state what is "given" before computing. 3 marks each
1.1 P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.2. Find P(A|B) and P(B|A).
1.2 A two-way table for 80 people shows 50 are female, 30 are male; 60 own a phone, 20 do not. Of the females, 40 own a phone. Build the table and find P(female | owns phone).
1.3 A die is rolled twice. Find P(sum > 10 | first roll is 6). (List the cases.)
1.4 A bag has 4 red and 6 blue marbles. Two marbles drawn without replacement. Find P(second red | first red).
1.5 Use Lesson 19's independence test (P(A|B) = P(A)?). Given P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.20, show whether A and B are independent. Show whether they are independent if instead P(A and B) = 0.15.
1.6 Year 10 class data: 18 boys, 12 girls. 8 boys and 6 girls play touch football. Find P(boy | plays touch football) and P(plays touch football | boy). Comment on which one is bigger and why.
2. Find the mistake
A Year 10 student tackles this: "In a class of 30 students, 18 play sport. Of those, 8 also play music. Find P(sport | music) given that 12 students play music in total." Exactly one line is wrong. 3 marks
Student's working:
Line 1: Given event is "music". Reduced sample = 12 students.
Line 2: Of those 12, 8 also play sport.
Line 3: P(sport | music) = 8 / 30 = 4/15.
Line 4: So P(sport | music) ≈ 0.267.
(a) Which line is wrong?
(b) Explain why, naming the Lesson 19 misconception.
(c) Write the corrected calculation.
Stuck? "Given" means divide by the SIZE of that given group — not by the whole class.3. Open-ended challenge — design your own "Simpson-style" scenario
Many valid answers. 4 marks
3.1 Design a real-world two-way table (rows = a categorical variable A with 2 categories, columns = a second categorical variable B with 2 categories) for exactly 200 people, where the conditional probabilities are different. Specifically, build the table so that:
- (i) P(B | A) ≠ P(B | not A) — i.e. A and B are NOT independent,
- (ii) at least one cell is at least 50,
- (iii) every cell is at least 5 (no empty corners).
Write up:
(a) the context in one sentence (e.g. "200 Year 10 students surveyed on…"),
(b) the completed two-way table with row/column totals,
(c) the two conditional probabilities P(B | A) and P(B | not A) with full working,
(d) a one-sentence interpretation in plain English ("knowing A makes B more / less likely because…").
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Formula both directions
P(A|B) = 0.2 / 0.4 = 0.5. P(B|A) = 0.2 / 0.5 = 0.4.
1.2 — Female | owns phone
50 female. 40 female own phone, so 10 female don't. 60 own phone, so 20 males own a phone. 30 male, so 10 male don't own a phone.
| Phone Y | Phone N | Total
Fem | 40 | 10 | 50
Male| 20 | 10 | 30
Tot | 60 | 20 | 80 ✓
P(female | owns phone) = 40 / 60 = 2/3.
1.3 — Sum > 10 given first is 6
Reduced sample: (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) → 6 outcomes. Sum > 10 in this set: (6,5) sum 11, (6,6) sum 12 → 2 outcomes.
P(sum > 10 | first = 6) = 2/6 = 1/3.
1.4 — Without replacement
After one red removed, 3 red remain in 9 marbles. P(second red | first red) = 3/9 = 1/3.
1.5 — Independence test (two cases)
Case 1: P(A|B) = 0.20/0.40 = 0.5 = P(A). Independent.
Case 2: P(A|B) = 0.15/0.40 = 0.375 ≠ P(A) = 0.5. Not independent.
1.6 — Touch football
Touch players = 8 + 6 = 14. P(boy | plays touch) = 8/14 = 4/7 ≈ 0.571. P(plays touch | boy) = 8/18 = 4/9 ≈ 0.444. P(boy | touch) is bigger because there are more boys than girls in touch (relative to their class share), and we are now dividing by the smaller "touch player" group rather than the larger "boys" group.
2 — Find the mistake
(a) Line 3.
(b) The student divided by 30 (the whole class) instead of 12 (the reduced sample of music players). Lesson 19 misconception fix: "P(A|B) means the probability of A given that B has occurred. The sample space is restricted to outcomes where B is true." So the denominator must be n(B) = 12, not the full 30.
(c) Corrected: P(sport | music) = 8 / 12 = 2/3 ≈ 0.667.
3 — Open-ended (sample solution)
Context: 200 Year 10 students surveyed on whether they own a bike (A) and whether they ride to school (B).
Table:
| Ride Y | Ride N | Total
Bike Y | 60 | 40 | 100
Bike N | 10 | 90 | 100
Total | 70 | 130 | 200 ✓
P(ride | own bike) = 60/100 = 0.60.
P(ride | no bike) = 10/100 = 0.10.
These are very different, so owning a bike is strongly associated with riding to school.
Interpretation: Knowing a student owns a bike makes them six times more likely to ride to school, because bike ownership is a near-prerequisite for riding.
Marking: 1 mark — context with totals to 200; 1 — valid table meeting all three rules; 1 — correct conditional probability calculations; 1 — sentence interpretation that connects to dependence.