PR-2: Solutions — Conditional Probability

Problem 1 — Reading Conditional Probability from a Two-Way Table

For every variant: (a) the denominator of is the marginal total for B (the conditioning event’s row/column total — not the grand total, not the intersection cell); (b) the conditional probability is the intersection cell over that marginal total.

Variant 0 — Study location × quiz result (n = 120):

	Passes	Does not pass	Total
Library	48	12	60
Elsewhere	36	24	60
Total	84	36	120

(a) Denominator = library row total = 60. (b) .

Variant 1 — Transportation × on-time (n = 200):

	On time	Late	Total
Public transit	72	48	120
Drives	56	24	80
Total	128	72	200

(a) Denominator = transit row total = 120. (b) .

Variant 2 — Membership × weekly frequency (n = 150):

	3+/week	Less frequent	Total
Student member	54	36	90
Non-student	24	36	60
Total	78	72	150

(a) Denominator = student row total = 90. (b) .

Variant 3 — Diet × energy level (n = 100):

	High energy	Low energy	Total
Vegetarian	28	12	40
Non-vegetarian	36	24	60
Total	64	36	100

(a) Denominator = vegetarian row total = 40. (b) .

Variant 4 — Work status × course completion (n = 200):

	Completes on time	Does not complete	Total
Works full-time	48	72	120
Not full-time	56	24	80
Total	104	96	200

(a) Denominator = full-time row total = 120. (b) .

Problem 2 — General Multiplication Rule (Generator)

Identify and , then apply .

Example: , → .

Common distractors: adding () gives a union-related value, not a joint; using one factor alone — both are required; reversing () goes backward (we have the conditional, want the joint, so multiply).

Problem 3 — Tree Diagram Path Probability (Generator)

Read the Stage-1 probability and the conditional Stage-2 probability from the branches, then multiply: .

Example: , → .

Common distractors: summing () — branch probabilities multiply along a path; wrong branch — follow the specified Stage-1 branch; forgetting Stage 1 — alone is conditional, multiply by .

Problem 4 — Independence Analysis

Data: 200 students; 90 play video games, 80 exercise, 36 both.

(a) .

(b) Multiplication test: , ; . The product equals the joint → independent (a surprising result — intuition might expect gamers to exercise less).

(c) Confirmation: — knowing a student exercises gives no information about gaming.

(d) Classmate’s claims: Claim 1 (“not mutually exclusive”) is correct (). Claim 2 (“co-occurrence proves independence”) is flawed — co-occurrence rules out mutual exclusivity but doesn’t imply independence; that requires the multiplication test. Here the test happens to confirm independence, but the reasoning is wrong.

Problem 1 — Two Conditional Probabilities from a Table

Compute both and : same numerator (intersection cell), different denominators. They are equal only when .

Variant 0 — Vitamins × colds (n = 200):

	Fewer colds	Normal/more	Total
Takes vitamins	72	48	120
No vitamins	36	44	80
Total	108	92	200

; . Not equal — same numerator (72), different denominators (120 vs. 108).

Variant 1 — Fiction × stress (n = 150):

	Lower stress	Normal/high	Total
Reads fiction	45	30	75
No fiction	30	45	75
Total	75	75	150

Both . Equal — special case only because .

Variant 2 — Bike commute × on-time (n = 200):

	On time	Late	Total
Bikes	56	24	80
Does not bike	72	48	120
Total	128	72	200

; . Not equal (denominators 80 vs. 128).

Variant 3 — Part-time × car ownership (n = 150):

	Owns car	No car	Total
Part-time	36	24	60
Not part-time	54	36	90
Total	90	60	150

; . Not equal. Note → these events are independent.

Variant 4 — Planner app × deadlines (n = 200):

	Meets all	Misses 1+	Total
Uses planner	60	40	100
No planner	40	60	100
Total	100	100	200

Both . Equal — special case ().

Problem 2 — Total Probability from a Tree Diagram (Generator)

Multiply along each path to the target outcome, then sum: .

Example: three suppliers with proportions 0.25, 0.35, 0.40 and defect rates 0.20, 0.10, 0.30.

Supplier	P(supplier)	P(defect ∣ supplier)	Joint
A	0.25	0.20	0.050
B	0.35	0.10	0.035
C	0.40	0.30	0.120

.

Problem 3 — Independence Test from Given Probabilities

Apply the multiplication test , then confirm via vs. .

• Variant 0 (coffee × car): → independent. Confirm: ✓.
• Variant 1 (news × votes): → dependent. Confirm: .
• Variant 2 (streaming × sports): → dependent. Confirm: .
• Variant 3 (exercise × sleep): → independent. Confirm: ✓.
• Variant 4 (budgeting × debt-free): → dependent. Confirm: .

Problem 4 — Find the Error

• Variant 0 — reversed conditioning: the doctor reported (sensitivity) but claimed it equals . Different denominators; for a rare disease, false positives can dwarf true positives, so can be far below 0.90. No numerical fix without the base rate.
• Variant 1 — wrong denominator: used instead of . Correct: (using yields the joint, not the conditional).
• Variant 2 — independence assumed: applied without testing. Test: → dependent; the correct joint is 0.25 (from the table).
• Variant 3 — assumed symmetry: computed then claimed “by symmetry.” They share a numerator but differ in denominator ( vs. ); correct (given). Compute each direction separately.
• Variant 4 — mutual exclusivity ≠ independence: concluded independence from . Backwards — , so the events are dependent (the most extreme form: B makes A impossible).

Problem 5 — Multi-Step Synthesis

Data: 500 customers; loyalty card 300/200; repeat purchase 180/320; both 150.

(a) Complete the table:

	Repeat	No repeat	Total
Loyalty card	150	150	300
No loyalty card	30	170	200
Total	180	320	500

Loyalty & no-repeat ; no-loyalty & repeat .

(b) Two conditionals: ; — loyalty holders repurchase at 3.3× the rate.

(c) Independence test: ; . → dependent.

(d) Reversed conditionals: vs. — not equal (denominators 180 vs. 300): asymmetry of conditioning.

(e) Manager vs. analyst: the manager correctly reports . The analyst raises valid points — observational data can’t establish causation (a confounder like income could drive both), and the relevant comparison is . Both are right: the arithmetic is correct; the causal interpretation is not warranted.

Problem 1 — Feynman Test

Conditioning direction specifies which event is known, which sets the denominator. restricts the universe to everyone who tested positive; restricts it to everyone with the disease — different denominators, different values.

Concrete example: prevalence 2%, sensitivity 0.90, specificity 0.95. Per 10,000 people: 200 have the disease; 180 true positives; 490 false positives; total positives = 670. Compare vs. — the disease is rare, so false positives (490) far outnumber true positives (180), diluting the positive predictive value.

Problem 2 — Two-Stage Quality Control

Stage 1 catches 85% of defectives; Stage 2 catches 70% of those reaching it; 4% of items are defective.

(a) A tree diagram is the right tool — sequential stages with conditional branch probabilities.

(b) For a defective item:

Stage 1	P	Stage 2	P (∣ passed S1)	Joint
Caught	0.85	—	—	0.85
Passes	0.15	Caught	0.70	0.105
Passes	0.15	Passes	0.30	0.045

, so .

(c) The stages are dependent: Stage 2 acts only on items that passed Stage 1, a restricted subpopulation. can’t equal an unconditional Stage-2 catch rate, because Stage 2 never sees items removed at Stage 1.

Problem 3 — Error Analysis

The student used because “there’s no reason these would be related.”

Error: applied the simplified multiplication rule without testing independence — “no reason to be related” is a contextual judgment, not a test.

Correct: (1) ; (2) compare to actual ; (3) → dependent. The correct joint is 0.20 (from data) — the product formula is valid only after independence is confirmed.

Path A — The Medical Analyst

Rare condition affecting 2%. Sensitivity ; specificity , so .

Task 1 — Tree (per 10,000):

Disease status	P	Test	P (cond)	Joint
Has disease (200)	0.02	Positive	0.90	0.018
Has disease (200)	0.02	Negative	0.10	0.002
No disease (9,800)	0.98	Positive	0.05	0.049
No disease (9,800)	0.98	Negative	0.95	0.931

Check: ✓.

Task 2 — : two paths → .

Task 3 — : . A 90%-sensitive test gives only ~27% probability that a positive reflects true disease, because the 9,800 healthy people generate 490 false positives, swamping the 180 true positives. Base rate dominates.

Task 4 — : , so . A negative result is very reassuring — only ~0.2% of negatives actually have the condition.

Key insight: for rare conditions, even accurate tests have high false-positive counts; positive predictive value depends critically on the base rate, which is why rare-disease screening uses confirmatory testing.

Path B — The Policy Analyst

5,000 records: young (≤25) = 1,500, older = 3,500. Accidents = 320 (180 young, 140 older). Company claims young-status and accident are independent.

Task 1 — Multiplication test: , ; product . . → dependent; the claim is false.

Task 2 — Conditional comparison: ; — young drivers crash at 3× the older rate. If independent, both would equal the overall 6.4%.

Task 3 — Consequence: using underestimates the true joint risk (0.036) by nearly 2×. In pricing, this under-collects premiums from the high-risk group, creating an actuarial shortfall subsidized by other policyholders.

Task 4 — Comparison: company’s vs. correct ; ratio — the company underestimates the joint probability by a factor of 1.875.

Challenge 1 — Algebraic Derivation

(a) General Multiplication Rule: from , multiply both sides by : ✓.

(b) Simplified rule: if independent, , so ✓.

(c) Independence in a tree: if the stages are independent, the Stage-2 branch probabilities are identical regardless of the Stage-1 branch — Stage 2 has “no memory” of Stage 1: .

Challenge 2 — Redundant Inspection Systems

Three independent inspectors each miss a defect with probability 0.15.

(a) .

(b) .

(c) Minimum inspectors for : gives (too high); gives ✓. Answer: 4 inspectors.

Challenge 3 — Conditional Probability Across Three Groups (Generator)

1. Within each group: conditional probability = intersection cell / that group’s row total.
2. Compare across groups to find the highest/lowest.
3. Reversed direction: uses the same cell over the outcome column total.

Key insight: and share the intersection cell but differ in denominator. The group with the highest forward rate is not necessarily the one that dominates among people with that outcome — that depends on group size.