Mathematics • Year 10 • Unit 4 • Lesson 19
Conditional Probability — Skill Drill
Build fluency with Lesson 19's core formula: P(A|B) = P(A and B) / P(B). Practise reading conditional probabilities from two-way tables (joint frequency / marginal total) and Venn diagrams (reduced sample space).
1. I do — fully worked example
Two ways to compute the same conditional probability — both should agree.
Problem. In a class of 30 students, 18 play sport and 12 play music. 8 students play both. Find P(music | sport) — "the probability a student plays music GIVEN that they play sport".
Step 1 — Identify the "given" condition.
Given B = "plays sport". This restricts the sample space to the 18 sport players.
Reason: Lesson 19 Key Term — "Reduced sample space: the subset of outcomes where the condition is satisfied."
Step 2 — Method 1: count from the reduced sample space.
Of the 18 sport players, 8 also play music.
P(music | sport) = 8 / 18 = 4/9 ≈ 0.444.
Reason: condition on B, then count A within that smaller pool.
Step 3 — Method 2: use the formula.
P(music and sport) = 8/30. P(sport) = 18/30.
P(music | sport) = (8/30) / (18/30) = 8/18 = 4/9. ✓
Reason: Lesson 19 Remember — P(A|B) = P(A and B) / P(B).
Step 4 — Sanity check the misconception.
P(A|B) is NOT P(A and B). 8/18 ≠ 8/30.
Answer: P(music | sport) = 4/9 ≈ 0.444.
2. We do — fill in the missing steps
Fill the blanks. 4 marks
Problem. The class above had 30 students, 18 play sport, 12 play music, 8 play both. Find P(sport | music).
Step 1 — Identify the "given".
Given B = "plays ____________". Reduced sample space = ____ students.
Step 2 — Method 1: reduced sample space.
Of the 12 music players, ____ also play sport.
P(sport | music) = ____ / ____ = ____ (fraction in simplest form).
Step 3 — Method 2: formula check.
P(sport and music) = 8/____ . P(music) = ____/____.
P(sport | music) = (8/30) ÷ (12/30) = ____.
Step 4 — Compare with P(music | sport).
P(music | sport) = 4/9 from Section 1. P(sport | music) = ____. They are __________ (equal / different) because the denominator (the "given") is different.
3. You do — independent practice
Show working. Foundation = one straight calc. Standard = two-way table reading. Extension = independence test using conditional probability.
Foundation — straight from the formula
3.1 P(A and B) = 0.15 and P(B) = 0.3. Find P(A|B). 1 mark
3.2 P(A and B) = 0.24 and P(A) = 0.6. Find P(B|A). 1 mark
3.3 A die is rolled. Find P(even | greater than 3). (Restrict to {4, 5, 6}; how many are even?) 1 mark
3.4 A card is drawn from a standard deck. Find P(heart | red card). 1 mark
Standard — read from a two-way table
3.5 A class of 50 students is surveyed for owning a phone (P) and owning a laptop (L). 30 own a phone, 25 own a laptop, 15 own both. Build the two-way table, then find P(laptop | phone) and P(phone | laptop). 3 marks
3.6 Apply Lesson 19's MCQ Q5 rule: "From a two-way table, conditional probability is found by dividing a joint frequency by a marginal total." Use this rule to find P(phone | NOT laptop) for the same class. 2 marks
Extension — independence via conditional probability
3.7 P(A) = 0.6, P(B) = 0.4, P(A and B) = 0.24. (a) Find P(A|B). (b) Use Lesson 19's misconception fix ("P(A|B) = P(A) when independent") to decide whether A and B are independent. 3 marks
3.8 A friend writes: "P(A|B) = 0 always, because if B has happened, A is no longer possible." Use the Lesson 19 misconception fix to write a one-sentence correction, and give one example where P(A|B) is between 0 and 1. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (P(sport | music))
Step 1: given B = "plays music"; reduced sample = 12.
Step 2: 8 of the 12 music players play sport. P(sport | music) = 8/12 = 2/3.
Step 3: 8/30 ÷ 12/30 = 8/12 = 2/3. ✓
Step 4: P(music | sport) = 4/9, P(sport | music) = 2/3. Different — the conditioning event changes the denominator.
3.1 — Direct formula
P(A|B) = 0.15 / 0.3 = 0.5.
3.2 — Direct formula
P(B|A) = 0.24 / 0.6 = 0.4.
3.3 — Die conditional
Reduced sample {4,5,6}. Even values: {4, 6} → 2 of 3. P(even | > 3) = 2/3.
3.4 — Heart | red card
26 red cards (13 hearts, 13 diamonds). P(heart | red) = 13/26 = 1/2.
3.5 — Phone and laptop table
Phone only = 15, both = 15, laptop only = 10, neither = 10.
| Lap Y | Lap N | Total
Ph Y | 15 | 15 | 30
Ph N | 10 | 10 | 20
Tot | 25 | 25 | 50 ✓
P(laptop | phone) = 15/30 = 1/2. P(phone | laptop) = 15/25 = 3/5.
3.6 — Phone | NOT laptop
"NOT laptop" column total = 25. Phone Y and NOT laptop = 15.
P(phone | NOT laptop) = 15/25 = 3/5.
3.7 — Independence via P(A|B)
(a) P(A|B) = 0.24 / 0.4 = 0.6.
(b) P(A|B) = 0.6 = P(A), so knowing B does not change the probability of A → A and B are independent.
3.8 — "P(A|B) = 0" misconception
Wrong. P(A|B) is the probability of A in the reduced sample space where B has happened — it can be anywhere between 0 and 1. Example: P(heart | red card) = 1/2, not 0.