Mathematics Advanced • Year 12 • Module 5 • Lesson 3
Conditional Probability
Build procedural fluency in the conditional formula, tree diagrams and two-way tables.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Write the conditional probability formula in three equivalent forms:
P(A | B) = _______________ / _______________ (with the restriction P(B) ______ 0)
P(A ∩ B) = P(B) × _______________ (multiplication rule rearranged)
P(A ∩ B) = P(A) × _______________ (symmetric form)
Q1.2 Explain in one sentence what "conditioning on B" does to the sample space.
Q1.3 A statement "P(A | B) = P(A)" means that A and B are ____________. Equivalently, P(A ∩ B) = _____________________.
2. Worked example — P(from X | defective)
Problem. A company sources components from two suppliers: 60% from X (4% defective) and 40% from Y (2% defective). A defective component is found. Find P(from X | defective).
Step 1 — Joint probabilities along each route.
P(X ∩ def) = P(X) × P(def | X) = 0.6 × 0.04 = 0.024
P(Y ∩ def) = P(Y) × P(def | Y) = 0.4 × 0.02 = 0.008
Reason: multiply along the branches of the tree (multiplication rule).
Step 2 — Total probability of being defective.
P(def) = 0.024 + 0.008 = 0.032
Reason: the law of total probability — sum across the partition {X, Y}.
Step 3 — Apply the conditional formula.
P(X | def) = P(X ∩ def) / P(def) = 0.024 / 0.032
= 24/32 = 3/4
Conclusion. Given the item is defective, P(it came from X) = 3/4 = 0.75. Notice this is higher than the unconditional P(X) = 0.6, because X has the higher defect rate — defective items disproportionately come from X.
3. Faded example — medical screening
A disease has prevalence P(D) = 0.01. A test has P(+ | D) = 0.99 (sensitivity) and P(+ | D′) = 0.05 (false-positive rate). Find P(D | +). 4 marks
Step 1 — Joint probabilities.
P(D ∩ +) = P(D) × P(+ | D) = ______ × ______ = ______
P(D′ ∩ +) = P(D′) × P(+ | D′) = ______ × ______ = ______
Step 2 — Total probability of a positive test.
P(+) = ______ + ______ = ______
Step 3 — Conditional formula.
P(D | +) = P(D ∩ +) / P(+) = ______ / ______ = ______
Conclusion. P(D | +) ≈ ______ (3 d.p.). The test is ______% accurate, yet a positive result gives only ______ probability of disease — because most positives come from the healthy majority.
4. Graduated practice
Foundation — direct formula application (4 questions)
| Q | Question | Working | Answer |
|---|---|---|---|
| 4.1 1 | P(A ∩ B) = 0.2, P(B) = 0.5. Find P(A | B). | ||
| 4.2 1 | P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(B | A). | ||
| 4.3 1 | P(A) = 0.6, P(B | A) = 0.3. Find P(A ∩ B). | ||
| 4.4 1 | If P(A | B) = P(A), what does this say about A and B? |
Standard — typical HSC difficulty (6 questions)
4.5 A bag has 5 red and 5 blue marbles. Two are drawn without replacement. Given the first is red, find P(second is red | first is red). 2 marks
4.6 In a class of 100 students: 30 have a passport, 20 have a passport and a driver's licence, and 60 have a driver's licence. Find P(passport | licence). 2 marks
4.7 A factory has three machines producing 50%, 30% and 20% of output, with defect rates 2%, 3% and 5%. Find the overall P(defective). 2 marks
4.8 From the table below, find P(work | sport). 2 marks
| Sport | No Sport | Total | |
|---|---|---|---|
| Work | 45 | 35 | 80 |
| No Work | 70 | 50 | 120 |
| Total | 115 | 85 | 200 |
4.9 Using the same table, find P(sport | work) and explain in one sentence why P(work | sport) ≠ P(sport | work). 2 marks
4.10 A drug test is 95% accurate (i.e. P(+ | user) = P(− | non-user) = 0.95). If 5% of athletes use the drug, find P(user | +) using a tree diagram. 3 marks
Extension — Bayes-style reasoning (2 questions)
4.11 Show algebraically that P(A | B) × P(B) = P(B | A) × P(A) for any two events with P(A), P(B) > 0. Hence derive Bayes' rule P(A | B) = P(B | A) × P(A) / P(B). 3 marks
4.12 Two events with P(A) = P(B) = 0.5 satisfy P(A | B) = 0.6. Find P(B | A) and P(A ∩ B), and decide whether A and B are independent. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Three forms
P(A | B) = P(A ∩ B) / P(B), P(B) ≠ 0. P(A ∩ B) = P(B) × P(A | B). P(A ∩ B) = P(A) × P(B | A).
Q1.2 — Meaning of conditioning
Conditioning on B restricts the sample space to outcomes where B has occurred; probabilities are then re-scaled so that within this smaller universe, the total probability is 1.
Q1.3 — Independence
A and B are independent; equivalently, P(A ∩ B) = P(A) × P(B).
Q3 — Faded example (disease test)
Step 1: P(D ∩ +) = 0.01 × 0.99 = 0.0099; P(D′ ∩ +) = 0.99 × 0.05 = 0.0495.
Step 2: P(+) = 0.0099 + 0.0495 = 0.0594.
Step 3: P(D | +) = 0.0099 / 0.0594 = 0.167 (3 d.p.) — about 1 in 6.
Conclusion: The test is 99% accurate (sensitivity), yet a positive result gives only ~17% probability of disease, because most positives are false positives drawn from the healthy 99% majority. This is the base-rate fallacy.
Q4.1 — Direct formula
P(A | B) = 0.2 / 0.5 = 0.4.
Q4.2 — Reverse direction
P(B | A) = P(A ∩ B) / P(A) = 0.2 / 0.4 = 0.5.
Q4.3 — Multiplication form
P(A ∩ B) = P(A) × P(B | A) = 0.6 × 0.3 = 0.18.
Q4.4 — Independence
P(A | B) = P(A) means A and B are independent — knowing B occurred gives no information about A.
Q4.5 — Marble draw
After one red is removed, 4 red and 5 blue remain (9 marbles total). P(second red | first red) = 4/9.
Q4.6 — Passport given licence
P(passport | licence) = P(passport ∩ licence) / P(licence) = 20/60 = 1/3.
Q4.7 — Total probability
P(def) = 0.50 × 0.02 + 0.30 × 0.03 + 0.20 × 0.05 = 0.010 + 0.009 + 0.010 = 0.029 (2.9%).
Q4.8 — Work | sport (from table)
P(work | sport) = 45/115 = 9/23 ≈ 0.391.
Q4.9 — Sport | work
P(sport | work) = 45/80 = 9/16 = 0.5625. The two are different because the conditioning event (and so the denominator) is different: in 4.8 we restrict to the 115 who play sport, in 4.9 we restrict to the 80 who work. Confusing them is the prosecutor's fallacy.
Q4.10 — Drug test
Tree: user (0.05) → + (0.95) or − (0.05); non-user (0.95) → + (0.05) or − (0.95).
P(+) = 0.05 × 0.95 + 0.95 × 0.05 = 0.0475 + 0.0475 = 0.095.
P(user | +) = 0.0475 / 0.095 = 0.5. Even at 95% accuracy, a positive result is only a coin-flip when the base rate is only 5%.
Q4.11 — Bayes derivation
By the multiplication rule, P(A ∩ B) = P(B) × P(A | B). By symmetry, P(A ∩ B) = P(A) × P(B | A). Equating:
P(A | B) × P(B) = P(B | A) × P(A), so dividing by P(B): P(A | B) = P(B | A) × P(A) / P(B). ▮
Q4.12 — Symmetric case
P(A ∩ B) = P(B) × P(A | B) = 0.5 × 0.6 = 0.3. P(B | A) = P(A ∩ B)/P(A) = 0.3/0.5 = 0.6. Check independence: P(A) × P(B) = 0.25 ≠ 0.3 = P(A ∩ B), so A and B are not independent (in fact they are positively associated).