L-PR-2: Solutions — Conditional Probability

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

The values in this section regenerate. Each solution therefore gives the method that works for every draw and one representative instance.

Problem 1 — Conditional Probability from a Table (C1 + C2)

Invariant method: in , first restrict the table to the conditioning event . The denominator is the marginal total for , and the numerator is the cell.

Representative instance:

	Passed	Did not pass	Total
Studies in a group	36	24	60
Studies alone	28	52	80
Total	64	76	140

Let be “passed” and be “studies in a group.”

Problem 2 — Testing Independence and Mutual Exclusivity (C5 + C6 + C7)

Invariant procedure:

1. Compute the observed joint probability using the overlap count divided by the grand total.
2. Compare it with . Equality means independent; a mismatch means dependent.
3. Confirm by comparing with .
4. Treat mutual exclusivity separately: a positive overlap rules it out, but does not prove independence.

Representative instance: among 200 students, 80 do activity , 100 do activity , and 50 do both.

Because , the events are dependent.

The conditional check also shows dependence. Since 50 students do both, the events are not mutually exclusive. Co-occurrence alone does not imply independence.

Problem 3 — Applying the General Multiplication Rule (C3)

Invariant method:

Representative instance: and .

The two factors describe reaching and then reaching within ; adding them would not produce the joint probability.

Problem 4 — Tree-Path Joint Probability (C3 + C4)

Invariant method: follow the requested path and multiply every branch probability along it.

Representative instance: a first-stage branch has probability , and the target second-stage outcome has conditional probability on that branch.

Problem 5 — Independence Test, Solo or Guided (C5 + C6)

Whether you choose the solo answer or the guided steps, use the same two checks.

Representative instance: in a group of 200 people, , , and 40 people do both, so .

The values match, so the events are independent.

The conditional probability equals the corresponding marginal probability, confirming the verdict.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Reversing Conditional Probabilities from a Table (C1 + C2 + C8)

Invariant method: the two directions share the same intersection numerator, but the event after the conditioning bar determines the denominator.

Representative instance:

	Fewer colds	Usual colds	Total
Takes vitamins	48	32	80
Does not take vitamins	36	84	120
Total	84	116	200

Let be “has fewer colds” and be “takes vitamins.”

They are not equal. Reversing the conditioning direction changes the denominator from 80 to 84.

Problem 2 — Total Probability across Tree Paths (C3 + C4)

Invariant method: multiply along every path that ends in the target outcome, then add those mutually exclusive path probabilities.

Representative instance: three suppliers provide , , and of inventory, with defect rates , , and .

Problem 3 — Independence by Product and Conditional Tests (C5 + C6)

Invariant method: compare the observed overlap with the product predicted by independence, then compute the requested conditional using the conditioning event’s count.

Representative instance: among 200 people, 80 are in , 100 are in , and 50 are in both.

The mismatch means the events are dependent.

The conditional test confirms the same verdict.

Problem 4 — Diagnosing Conditional-Probability Errors (C1 + C2 + C5 + C6 + C7 + C8)

The generator selects one of five error diagnoses. Use the correction matching the case you received:

1. Reversed conditioning: is , not . The reversed conditional also needs the disease prevalence.

2. Wrong denominator: if and , then

   Dividing by merely returns the joint probability.

3. Unverified independence: if , , and the observed joint probability is , then

   Since , the events are dependent; the table’s is the correct joint probability.

4. False symmetry: does not imply . In the generated case the table gives ; the denominators differ.

5. Mutual exclusivity confused with independence: for positive-probability events, implies

   Since , , so the events are dependent.

Problem 5 — Loyalty-Card Probability Synthesis (C1 + C2 + C5 + C6 + C8)

This item is fixed: 500 customers, 300 with a loyalty card, 180 repeat purchasers, and 150 who are both loyalty-card holders and repeat purchasers.

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

Complete the table:

Compute the two forward conditionals:

Test independence:

Because , repeat purchase and loyalty-card status are dependent.

Reverse the conditioning direction:

This differs from because the denominator changes.

Interpretation: both speakers make a valid point. The manager’s 50% conditional probability is arithmetically correct, but observational data do not show that the loyalty program caused the repeat purchase. The 15% comparison rate for non-cardholders is also relevant.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

The conditioning direction specifies what is already known and therefore which group forms the denominator. restricts attention to people who tested positive; restricts attention to people who have the disease.

For a disease with 2% prevalence, 90% sensitivity, and 95% specificity, consider 10,000 people.

Thus and differ by about . The many people without the rare disease can produce more false positives than the diseased group produces true positives.

Question 2 — Two-Stage Quality Control

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

(a) A tree diagram is appropriate because the inspections are sequential and the stage-2 probabilities are conditional on reaching stage 2.

(b) A defective item must pass both inspections.

(c) The stage events are dependent by design. Stage 2 only acts on items that passed stage 1, so it operates on a restricted population rather than on every defective item.

Question 3 — Error Analysis

The student used without establishing independence.

Since , the events are dependent. The correct joint probability is the observed value ; contextual intuition cannot replace the independence test.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Both Boss workspaces regenerate. Each has four integrated tasks and an optional solo route; the method below applies to every draw.

Path A — The Medical Analyst

Invariant method: build the positive and negative paths, add paths ending in the same test result, and use the relevant result total as the denominator of the reversed conditional.

Representative instance: prevalence , sensitivity , and specificity . Therefore the false-positive rate is and the false-negative rate is .

Task 1 — Find the total positive-test probability.

Task 2 — Reverse the conditioning direction for a positive result.

Task 3 — Analyze a negative result.

Task 4 — Explain the result. Sensitivity is , not . Because the base rate is only 2%, the much larger no-condition group can generate more false positives than the condition group generates true positives. A negative result reduces the probability sharply because condition-and-negative cases are rare.

Model reflection: The hardest part was attaching each conditioning direction to the correct denominator. Next time, I will write the target conditional in words, calculate the required joint paths, and check that the denominator represents the group named after “given.”

Path B — The Policy Analyst

Invariant method: compute the joint probability predicted by independence, compute the observed joint probability from the table, compare them and quantify the ratio, then explain the decision consequence without claiming causation.

Representative instance: 5,000 insurance records include 1,500 young drivers, 320 drivers with an accident, and 180 young drivers with an accident.

Task 1 — Compute the company’s independence-based prediction.

Task 2 — Compute the observed joint probability.

Task 3 — Diagnose and quantify the mismatch.

The events are dependent.

Task 4 — Explain the practical consequence. The observed co-occurrence is about times the independence-based prediction, so a decision model using the smaller prediction would systematically underestimate the observed joint rate for this group. More data can refine the estimates, but it cannot repair an invalid independence assumption. This association identifies dependence, not its cause.

Model reflection: The observed joint probability changed my judgment because it did not match the product predicted by independence. Before recommending a policy change, I would verify the table’s definitions and data quality, then investigate group differences or selection mechanisms that might explain the association.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Algebraic Derivation

(a) Derive the General Multiplication Rule.

Assume , so the conditional probability below is defined.

(b) Apply independence. If and are independent, then because .

(c) Interpret independence in a tree. For stage-1 branches with positive probability, the stage-2 probabilities are the same after every branch. In particular, if and ,

Stage 2 has no statistical “memory” of the stage-1 outcome.

Challenge 2 — Redundant Inspection Systems

Each of the three generator blocks draws its own miss probability . Read the displayed for that block rather than carrying a value from another part.

(a) Invariant method:

For a representative draw with :

(b) Invariant method:

For a representative draw with :

(c) Invariant method: find the smallest integer satisfying . Check both and to prove minimality.

For a representative draw with :

The minimum is 4 inspectors.

Challenge 3 — Conditional Probability Across Three Groups

Invariant method: divide each outcome count by its group row total, compare those forward conditionals, then reverse the direction by dividing the first group’s outcome count by the outcome column total.

Representative instance:

Age group	Adopts new technology	Does not adopt	Total
18–30	10	40	50
31–50	60	40	100
51+	20	30	50
Total	90	110	200

The 31–50 group has the highest conditional adoption probability.

This reversed conditional is not the forward rate : the numerator is unchanged, but the denominator is now all adopters.