PR-2 Solutions: Conditional Probability

IP1 — Two Conditional Probabilities from a Table

For every variant: compute both \( P(A \mid B) \) and \( P(B \mid A) \) separately. They share the same numerator (the intersection cell) but different denominators. They are generally not equal — only when \( P(A) = P(B) \).

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

• \( P(\text{fewer colds} \mid \text{vitamins}) = 72/120 = \mathbf{0.60} \)
• \( P(\text{vitamins} \mid \text{fewer colds}) = 72/108 \approx \mathbf{0.667} \)
• Not equal — same numerator (72), different denominators (120 vs. 108). Direction matters.

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

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

• \( P(\text{lower stress} \mid \text{fiction}) = 45/75 = \mathbf{0.60} \)
• \( P(\text{fiction} \mid \text{lower stress}) = 45/75 = \mathbf{0.60} \)
• Equal — but this is a special case because both row and column marginals for the two events both equal 75. In general, \( P(A \mid B) \neq P(B \mid A) \). The equality here holds only because \( P(A) = P(B) = 0.50 \).

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

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

• \( P(\text{on time} \mid \text{bikes}) = 56/80 = \mathbf{0.70} \)
• \( P(\text{bikes} \mid \text{on time}) = 56/128 = \mathbf{0.4375} \)
• Not equal — 0.70 ≠ 0.4375. The denominators differ: 80 (bike commuters) vs. 128 (on-time arrivals).

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

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

• \( P(\text{car} \mid \text{part-time}) = 36/60 = \mathbf{0.60} \)
• \( P(\text{part-time} \mid \text{car}) = 36/90 = \mathbf{0.40} \)
• Not equal — 0.60 ≠ 0.40. Note: \( P(\text{car} \mid \text{part-time}) = 0.60 = P(\text{car}) = 90/150 = 0.60 \), confirming these events are independent.

Variant 4 — Planner app × deadline adherence (n = 200)

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

• \( P(\text{meets} \mid \text{planner}) = 60/100 = \mathbf{0.60} \)
• \( P(\text{planner} \mid \text{meets}) = 60/100 = \mathbf{0.60} \)
• Equal — special case because both marginals equal 100 (i.e., \( P(\text{planner}) = P(\text{meets}) = 0.50 \)). The equality holds only when \( P(A) = P(B) \), not in general.

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IP2 — Total Probability from a Tree Diagram (Generator)

Generated by generateTotalProbabilityProblem(). The approach for every instance:

1. Identify the target Stage-2 outcome (e.g., "defective").
2. Multiply along each path that leads to that outcome: \( P(\text{path}) = P(\text{Stage-1 branch}) \times P(\text{Stage-2 branch} \mid \text{Stage-1}) \).
3. Sum all such paths: \( P(\text{target}) = \sum_i P(\text{path}_i) \).

Example: Three suppliers A, B, C 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	0.25 × 0.20 = 0.050
B	0.35	0.10	0.35 × 0.10 = 0.035
C	0.40	0.30	0.40 × 0.30 = 0.120

\[ P(\text{defective}) = 0.050 + 0.035 + 0.120 = 0.205 \]

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IP3 — Independence Test from Given Probabilities

For every variant: apply the multiplication test \( P(A) \cdot P(B) \overset{?}{=} P(A \cap B) \). Then confirm by computing \( P(A \mid B) = P(A \cap B) / P(B) \) and comparing to \( P(A) \).

Variant 0 — Premium coffee × car ownership

• \( P(\text{coffee}) \cdot P(\text{car}) = 0.40 \times 0.55 = 0.22 = P(\text{both}) \) → INDEPENDENT
• Confirm: \( P(\text{coffee} \mid \text{car}) = 0.22 / 0.55 = 0.40 = P(\text{coffee}) \) ✓

Variant 1 — Reads news × votes

• \( 0.35 \times 0.60 = 0.21 \neq 0.27 = P(\text{both}) \) → DEPENDENT
• Confirm: \( P(\text{news} \mid \text{votes}) = 0.27 / 0.60 = 0.45 \neq 0.35 = P(\text{news}) \)

Variant 2 — Streaming × live sports

• \( 0.70 \times 0.45 = 0.315 \neq 0.28 = P(\text{both}) \) → DEPENDENT
• Confirm: \( P(\text{sports} \mid \text{streams}) = 0.28 / 0.70 = 0.40 \neq 0.45 = P(\text{sports}) \)

Variant 3 — Exercises 3+/week × sleeps 7+ hours

• \( 0.30 \times 0.50 = 0.15 = P(\text{both}) \) → INDEPENDENT
• Confirm: \( P(\text{exercise} \mid \text{sleep well}) = 0.15 / 0.50 = 0.30 = P(\text{exercise}) \) ✓

Variant 4 — Budgeting app × debt-free

• \( 0.25 \times 0.60 = 0.15 \neq 0.18 = P(\text{both}) \) → DEPENDENT
• Confirm: \( P(\text{debt-free} \mid \text{budgets}) = 0.18 / 0.25 = 0.72 \neq 0.60 = P(\text{debt-free}) \)

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IP4 — Find the Error

Variant 0 — Doctor's claim (reversed conditioning)

Error: The doctor stated \( P(\text{positive} \mid \text{disease}) = 0.90 \) (sensitivity) but claimed it equals \( P(\text{disease} \mid \text{positive}) \). These two conditionals have different denominators — one restricts to diseased patients, the other to patients who tested positive. For a rare disease, the false-positive pool can dwarf the true-positive pool, making \( P(\text{disease} \mid \text{positive}) \) dramatically lower than 0.90 even with a highly sensitive test.

No numerical correction is possible without the base rate (prevalence).

Variant 1 — Wrong denominator

Error: The student used \( P(S) = 1 \) as the denominator instead of \( P(B) = 0.40 \).

Correct calculation:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.40} = 0.30 \]

Using \( P(S) = 1 \) as denominator gives \( P(A \cap B) \), not \( P(A \mid B) \).

Variant 2 — Independence assumed without testing

Error: The student applied the simplified multiplication rule \( P(A \cap B) = P(A) \cdot P(B) \) without first verifying independence.

Independence test: \( P(\text{on time}) \cdot P(\text{bus}) = 0.65 \times 0.30 = 0.195 \neq 0.25 = P(\text{on time} \cap \text{bus}) \).

The events are dependent. The correct joint probability is 0.25 (from the table), not 0.195.

Variant 3 — Assumed symmetry of conditioning

Error: The student correctly computed \( P(B \mid A) = 0.42 \) but then claimed \( P(A \mid B) = 0.42 \) "by symmetry."

The two conditionals share the numerator \( P(A \cap B) \) but have different denominators — \( P(B) \) for \( P(A \mid B) \) and \( P(A) \) for \( P(B \mid A) \). They are equal only when \( P(A) = P(B) \).

Correct result: \( P(A \mid B) = 0.28 \) (given in the problem). Always compute each direction separately.

Variant 4 — Mutual exclusivity confused with independence

Error: The student concluded independence from \( P(A \cap B) = 0 \) (mutual exclusivity). This is backwards — mutually exclusive events with positive probability are always dependent.

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{P(B)} = 0 \neq P(A) \]

Since \( P(A \mid B) = 0 \neq P(A) \), the events are dependent. Knowing \( B \) occurred makes \( A \) impossible — the most extreme form of dependence, not independence.

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IP5 — Multi-Step Synthesis (non-regenerable)

Data: 500 customers. Loyalty card: 300 yes, 200 no. Repeat purchase: 180 yes, 320 no. Both (loyalty AND repeat): 150.

Part (a) — Complete the 2×2 table:

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

Loyalty card, no repeat = 300 − 150 = 150. No loyalty, repeat = 180 − 150 = 30.

Part (b) — Two conditional probabilities:

\[ P(\text{repeat} \mid \text{loyalty}) = \frac{150}{300} = 0.50 \] \[ P(\text{repeat} \mid \text{no loyalty}) = \frac{30}{200} = 0.15 \]

These differ substantially — loyalty card holders repurchase at 3.3× the rate of non-holders.

Part (c) — Independence test:

\[ P(\text{repeat}) \cdot P(\text{loyalty}) = \frac{180}{500} \times \frac{300}{500} = 0.36 \times 0.60 = 0.216 \] \[ P(\text{repeat} \cap \text{loyalty}) = \frac{150}{500} = 0.30 \]

\( 0.216 \neq 0.30 \) → the events are DEPENDENT.

Part (d) — Reversed conditional probabilities:

\[ P(\text{loyalty} \mid \text{repeat}) = \frac{150}{180} \approx 0.833 \] \[ P(\text{repeat} \mid \text{loyalty}) = \frac{150}{300} = 0.50 \]

These are not equal (0.833 ≠ 0.50). They share numerator 150 but use different denominators (180 repeat purchasers vs. 300 loyalty card holders). This is a concrete example of C8 — asymmetry of conditioning.

Part (e) — Evaluating the manager vs. analyst:

• The manager correctly states that \( P(\text{repeat} \mid \text{loyalty}) = 0.50 \). The 50% figure is the right conditional probability to report for loyalty card holders.
• The analyst raises valid methodological concerns: (1) observational data cannot establish causation — a confounding variable (e.g., higher income) could explain both loyalty card adoption and repeat purchasing; (2) the relevant comparison is to \( P(\text{repeat} \mid \text{no loyalty}) = 0.15 \), which contextualizes the 50%.
• Both make valid points. The arithmetic is correct; the causal interpretation is not warranted.

Question 1 — Feynman Test (Model Answer)

See the model answer block in the lesson for the full explanation. The key ideas:

• Conditioning direction specifies which event we know has occurred — it determines the denominator. \( P(\text{disease} \mid \text{positive}) \) restricts the universe to everyone who tested positive; \( P(\text{positive} \mid \text{disease}) \) restricts it to everyone who has the disease. Different denominators, different values.
• Concrete example: Disease prevalence 2%, sensitivity 0.90, specificity 0.95. Per 10,000 people: 200 have the disease; 180 test positive (true positives); 490 test positive falsely (false positives). Total positive tests = 670.

\[ P(\text{disease} \mid \text{positive}) = \frac{180}{670} \approx 0.27 \]

Compare: \( P(\text{positive} \mid \text{disease}) = 0.90 \) vs. \( P(\text{disease} \mid \text{positive}) \approx 0.27 \). The difference is 0.63 — far greater than 0.50. The reason: disease is rare (base rate 2%), so false positives (490) far outnumber true positives (180), diluting the positive predictive value.

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Question 2 — Two-Stage Quality Control

Stage 1 catches 85% of defectives. Stage 2 catches 70% of the defectives that reach it. 4% of items are defective.

(a) A tree diagram is the most appropriate tool — sequential stages with conditional probabilities at each branch. Answer: tree diagram.

(b) Tree for a defective item:

Stage 1	P	Stage 2	P (given passed S1)	Joint (defective path)
Caught	0.85	—	—	0.85
Passes	0.15	Caught	0.70	0.15 × 0.70 = 0.105
Passes	0.15	Passes	0.30	0.15 × 0.30 = 0.045

\( P(\text{passes both} \mid \text{defective}) = 0.15 \times 0.30 = 0.045 \)

\[ P(\text{defective AND passes both}) = P(\text{defective}) \times P(\text{passes both} \mid \text{defective}) = 0.04 \times 0.045 = \mathbf{0.0018} \]

(c) The two stages are dependent. Stage 2 only acts on items that passed Stage 1 — it operates on a restricted subpopulation. The conditional probability \( P(\text{caught at Stage 2} \mid \text{defective, passed Stage 1}) = 0.70 \) cannot equal the unconditional probability of being caught at Stage 2 across all defectives, because Stage 2 never sees items removed at Stage 1.

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Question 3 — Error Analysis

The student applied \( P(A \cap B) = P(A) \cdot P(B) = 0.35 \times 0.40 = 0.14 \) because "there's no reason these would be related."

Specific error: Applied the simplified multiplication rule without first testing independence. "No reason to be related" is a contextual judgment, not a mathematical test.

Correct procedure:

1. Compute \( P(A) \cdot P(B) = 0.35 \times 0.40 = 0.14 \).
2. Compare to the actual \( P(A \cap B) = 0.20 \) (from data).
3. Since \( 0.14 \neq 0.20 \), the events are dependent.

The correct joint probability is 0.20 — read from the data. The formula \( P(A) \cdot P(B) \) is only valid after independence is confirmed; it cannot replace actual data.

Path A — The Medical Analyst

Scenario: rare condition affecting 2% of the population. Sensitivity = \( P(\text{positive} \mid \text{disease}) = 0.90 \). Specificity = \( P(\text{negative} \mid \text{no disease}) = 0.95 \), so \( P(\text{positive} \mid \text{no disease}) = 0.05 \).

Task 1 — Complete tree (per 10,000 people):

Stage 1 (Disease status)	P	Stage 2 (Test result)	P (conditional)	Joint probability
Has disease (200 people)	0.02	Positive	0.90	0.02 × 0.90 = 0.018
Has disease (200 people)	0.02	Negative	0.10	0.02 × 0.10 = 0.002
No disease (9,800 people)	0.98	Positive	0.05	0.98 × 0.05 = 0.049
No disease (9,800 people)	0.98	Negative	0.95	0.98 × 0.95 = 0.931

Check: 0.018 + 0.002 + 0.049 + 0.931 = 1.000 ✓

Task 2 — \( P(\text{tests positive}) \):

Two paths lead to a positive test: (has disease AND positive) and (no disease AND positive).

\[ P(\text{positive}) = 0.018 + 0.049 = 0.067 \]

Task 3 — \( P(\text{disease} \mid \text{positive}) \):

\[ P(\text{disease} \mid \text{positive}) = \frac{P(\text{disease} \cap \text{positive})}{P(\text{positive})} = \frac{0.018}{0.067} \approx \mathbf{0.269} \]

Compare to the 90% sensitivity: a test with 90% sensitivity gives only ~27% probability that a positive result reflects true disease. This is surprising because the disease is rare (2% base rate) — even with low false-positive rate (5%), the 9,800 healthy people generate 490 false positives, which swamp the 180 true positives (200 × 0.90). Base rate dominates.

Task 4 — \( P(\text{disease} \mid \text{negative}) \):

\[ P(\text{negative}) = 0.002 + 0.931 = 0.933 \] \[ P(\text{disease} \mid \text{negative}) = \frac{0.002}{0.933} \approx 0.0021 \]

A negative test result is very reassuring — only about 0.2% of people who test negative actually have the condition. The test's 90% sensitivity means it misses 10% of true cases (20 out of 200), but that is a small absolute number compared to the 9,310 true negatives.

Key insight: For rare conditions, even accurate tests have high false-positive rates in absolute terms. Positive predictive value (\( P(\text{disease} \mid \text{positive}) \)) depends critically on the base rate — screening programs for rare diseases typically require confirmatory testing precisely for this reason.

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Path B — The Policy Analyst

Scenario: 5,000 records. Young drivers (≤25): 1,500. Older drivers (>25): 3,500. Accidents: 320 total (180 young, 140 older). Company claims young-driver status and accident are independent.

Task 1 — Multiplication test:

\[ P(\text{young}) = \frac{1500}{5000} = 0.30, \quad P(\text{accident}) = \frac{320}{5000} = 0.064 \] \[ P(\text{young}) \cdot P(\text{accident}) = 0.30 \times 0.064 = 0.0192 \] \[ P(\text{young} \cap \text{accident}) = \frac{180}{5000} = 0.036 \]

\( 0.0192 \neq 0.036 \) → the events are DEPENDENT. The independence claim is false.

Task 2 — Conditional probability comparison:

\[ P(\text{accident} \mid \text{young}) = \frac{180}{1500} = 0.120 \] \[ P(\text{accident} \mid \text{older}) = \frac{140}{3500} = 0.040 \]

The accident rate for young drivers (12.0%) is three times that for older drivers (4.0%). These conditional probabilities are far from equal — confirming dependence. If the events were independent, both rates would equal the overall rate (6.4%).

Task 3 — Consequence of incorrect independence assumption:

When the company applies \( P(\text{young} \cap \text{accident}) = P(\text{young}) \cdot P(\text{accident}) = 0.0192 \), it underestimates the true joint risk (0.036) by a factor of nearly 2. In insurance pricing: underestimating risk means collecting insufficient premiums from the high-risk group (young drivers), which creates an actuarial shortfall. The company effectively subsidizes young-driver risk through other policyholders.

Task 4 — Comparison of correct vs. incorrect joint probability:

• Company's calculation: \( P(\text{young} \cap \text{accident}) = 0.0192 \)
• Correct value from table: \( P(\text{young} \cap \text{accident}) = 0.036 \)
• Ratio: \( 0.036 / 0.0192 = 1.875 \) — the company underestimates the joint probability by a factor of 1.875 (approximately 88% underestimate).

Challenge 1 — Algebraic Derivation

Part (a) — Derive the General Multiplication Rule:

Start from the conditional probability formula:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]

Multiply both sides by \( P(B) \):

\[ P(B) \cdot P(A \mid B) = P(A \cap B) \]

This is the General Multiplication Rule: \( P(A \cap B) = P(B) \cdot P(A \mid B) \). ✓

Part (b) — Derive the Simplified Multiplication Rule:

If \( A \) and \( B \) are independent, then by definition \( P(A \mid B) = P(A) \). Substitute into the General Rule:

\[ P(A \cap B) = P(B) \cdot P(A \mid B) = P(B) \cdot P(A) = P(A) \cdot P(B) \]

This is the Simplified Multiplication Rule for independent events: \( P(A \cap B) = P(A) \cdot P(B) \). ✓

Part (c) — Independence in a tree diagram:

If Stage 1 and Stage 2 are independent, the conditional probabilities on Stage 2 branches are the same regardless of which Stage 1 branch was taken. Stage 2 has "no memory" of Stage 1 — the branch probabilities at each Stage 2 node are identical. Formally: \( P(B \mid A) = P(B \mid A') = P(B) \) for any Stage 2 event \( B \).

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Challenge 2 — Redundant Inspection Systems

Each of three independent inspectors misses a defect with probability 0.15.

Part (a) — P(all three miss):

By independence, multiply the individual miss probabilities:

\[ P(\text{all miss}) = 0.15 \times 0.15 \times 0.15 = (0.15)^3 = 0.003375 \]

Part (b) — P(at least one catches):

Use the complement rule — "at least one catches" is the complement of "all miss":

\[ P(\text{at least one catches}) = 1 - P(\text{all miss}) = 1 - 0.003375 = \mathbf{0.996625} \]

Part (c) — Minimum inspectors to get P(all miss) < 0.001:

• \( n = 3 \): \( (0.15)^3 = 0.003375 \) — exceeds 0.001. Not sufficient.
• \( n = 4 \): \( (0.15)^4 = 0.00050625 \approx 0.000506 \) — below 0.001. ✓

Answer: 4 inspectors. The minimum to achieve \( P(\text{all miss}) \lt 0.001 \) is \( n = 4 \).

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Challenge 3 — Conditional Probability Across Three Groups (Generator)

Generated by generateThreeGroupCondProbProblem(). The approach for any instance:

1. Within each group: compute the conditional probability as (intersection cell) / (row total for that group).
2. Compare across groups: identify which group has the highest or lowest conditional probability.
3. Reversed direction: to find \( P(\text{group} \mid \text{outcome}) \), use the same intersection cell but divide by the column total for the outcome.

The key insight: \( P(\text{outcome} \mid \text{group}) \) and \( P(\text{group} \mid \text{outcome}) \) use the same intersection cell but different denominators. The group with the highest forward conditional rate is not necessarily the group that "dominates" among people with that outcome — that depends on group size.