L-PR-2: Conditional Probability

Module 2 · Probability Foundations

Section 1: Introduction

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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.

A rapid COVID test shows a positive result. The test has 95% sensitivity and 95% specificity. Does that mean there’s a 95% chance you actually have COVID?

This is a conditional probability question — and the answer might surprise you. At 1% prevalence, 95% sensitivity, and 95% specificity, the chance of actually having the disease after a positive result is only about 16.1%. The mathematics behind this apparent paradox is the subject of today’s lesson.

In L-PR-1, you learned to compute — the probability that both events occur. That joint probability is a building block. In L-PR-2, we use it as the starting point for a more nuanced question: given that we already know one thing has happened, how does that change the probability of something else?

After this lesson, you will be able to:

Compute conditional probabilities using the formula and directly from two-way frequency tables.
Apply the General Multiplication Rule to find joint probabilities.
Construct and use tree diagrams to compute probabilities in multi-stage experiments.
Test whether two events are independent using both equivalent definitions.

In L-PR-1, you also saw a preview: mutually exclusive events and independent events are not the same thing. By the end of this lesson, you will have the tools to state precisely why — and to prove it algebraically.

Section 2: Prerequisites

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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.

What you need from L-PR-1

Reading joint and marginal frequencies from two-way tables (L-PR-1, Core Concepts C7–C9): Today’s conditional probability formula has a numerator and a denominator. Both come from table cells. You need to identify the joint count (intersection cell) and the marginal total (row or column total) quickly and reliably.
Computing (L-PR-1, Core Concepts C8): The joint probability is the numerator of the conditional probability formula. If you can read it from a table, you already have half the formula.
Identifying mutually exclusive events (L-PR-1, Core Concepts C10): L-PR-1 ended with a preview: mutually exclusive events and independent events are distinct concepts. Today we resolve that preview formally — you need to know what “mutually exclusive” means to appreciate why it implies dependence.

Retrieval Checkpoint

A school surveyed 80 students about study habits and exam results. The two-way table below shows the data.

	Passed exam	Did not pass	Total
Studies in a group	36	9	45
Studies alone	24	11	35
Total	60	20	80

Also confirm you can recall:

I can distinguish the intersection cell from a row total or column total in a two-way table. I know that

— intersection is symmetric. I can identify when two events are mutually exclusive (

Success Factor:

From endpoint to numerator. In L-PR-1, was an endpoint — the answer to “what is the probability of both?” In L-PR-2, it becomes the numerator of a more nuanced calculation. We also resolve the mystery from L-PR-1 C10: what does “independent” mean precisely, and how is it different from “mutually exclusive”?

Section 3: Core Concepts

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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.

Navigation Guide — 8 Concepts

C1–C2: The core idea — restricting the sample space — and the formula that captures it. Use these every time you see “given.”
C3–C4: Two tools for computing joint probabilities: rearranging the formula (Multiplication Rule) and drawing tree diagrams.
C5–C6: What independence means precisely, and how to test it — two equivalent ways.
C7: Resolves the L-PR-1 preview: why mutually exclusive events with positive probability are dependent.
C8: The most dangerous mistake in applied probability — reversing the conditioning direction.

C1 — Conditional Probability: The Concept

Imagine a bag with 20 marbles: 8 red, 6 blue, 6 green. If you draw a marble at random, .

Now suppose someone peeks and tells you: “The marble is not green.” That information restricts the sample space. You no longer consider all 20 marbles — only the 14 that are not green. Of those 14, how many are red? Still 8. So the probability of red, given not green, is .

This is the idea of conditional probability. Conditioning on an event means:

Throw away all outcomes not in — they are impossible given what we know.
Renormalize — the remaining outcomes now form the new universe, and their probabilities must sum to 1.

We write — read “the probability of given ” — for the probability of once the sample space has been restricted to .

Conditional Probability — Restricted Sample Space

The conditional probability is the probability of after restricting the sample space to the outcomes in , provided . Within that restricted sample space, the outcomes in are the outcomes favourable to .

See it with the marble bag: Click “Condition on ‘not green’” to throw away the green marbles. Watch the universe shrink from 20 to 14 while the 8 red marbles stay — so rises from to . The denominator shrank; the numerator did not.

Universe size: 20 marbles

P(red) = 8 / 20 = 0.40

Figure 1: Conditioning restricts the sample space. Removing the six green marbles leaves the eight red marbles in a smaller universe, so the denominator changes from 20 to 14 and the probability of red rises.

When conditioning on , the denominator is — not . Using the full sample space as denominator gives , not . Every time you see the word “given,” ask: what is the restricting event? That is always the denominator.

C2 — Conditional Probability: The Formula

Conditional Probability Formula

For any events and with :

The numerator is the joint probability of both events. The denominator is the probability of the conditioning event (the event we know has occurred).

Reading conditional probabilities directly from a two-way table:

When data is presented as counts in a table, you don’t need to compute and separately. Work directly with counts:

The numerator is the cell where row and column (or row and column ) intersect. The denominator is the row or column total for — not the grand total.

Mini-example — two-way table method:

In the retrieval checkpoint table above (80 students): what is ?

Numerator: 36 students who both study in a group and passed. Denominator: 45 students who study in a group (row total for ).

Note: this is NOT the same as . Direction matters.

and are computed with the same numerator () but different denominators ( vs. ). They are equal only in special cases. Never swap the conditioning direction — it changes the meaning of the calculation entirely.

See the denominator switch in action:

Click “Condition on row B” to collapse the non-B rows — the denominator moves from the grand total to B’s row total. Then click “Condition on column A” instead: the numerator (the cell) stays the same, but now the denominator is A’s column total. Same numerator, different denominator — that is exactly why (a preview of the asymmetry in C8).

	A (Passed exam)	A′ (Did not pass)	Row total
B (Studies in group)	36	9	45
B′ (Studies alone)	24	11	35
Column total	60	20	80

P(A | B) = count(A ∩ B) = 36 denominator = 80 = 36/80 = 0.45 ← P(A∩B), not a conditional

Dividing by n = 80 gives P(A∩B) = 0.45, not a conditional probability. Pick a conditioning event to restrict the universe.

Same numerator, different denominator: P(A | B) = 36/45 = 0.80 ≠ P(B | A) = 36/60 = 0.60. Direction matters.

Each mode keeps the same A∩B cell but changes the denominator

Figure 2: Conditioning keeps the same intersection count, 36, but changes the reference group. Restricting to row B uses 45 as the denominator, while restricting to column A uses 60, so P(A | B) and P(B | A) are different.

Now see the same idea geometrically:

Where the table above shows the denominator switching between counts, the mosaic below shows it as area. The sample space is a square; B is the left column (its width is ) and A is the top band of each column (its height is the conditional probability of A in that column). So is literally how tall the A-slice is inside B’s column.

P(A) = 0.30

P(B) = 0.25

P(A∩B) = 0.1000

P(A∩B) stays within the bounds supported by P(A) and P(B).

P(A) = 0.3000 P(B) = 0.25 P(A∩B) = 0.1000

P(A | B) = P(A∩B) / P(B) = 0.1000 / 0.25 = 0.4000

Figure 3: The mosaic links probability area, intersections, and conditional probability. Adjust the event sizes and overlap, then condition on B to see P(A | B) as the share of B that also belongs to A.

Drag the sliders to resize A, B, and their overlap, then click “Condition on B” to crop the square down to B’s column. Notice how changes when B is wide vs. narrow — even for the same overlap. Watch the dashed A-line: when it sits at the same height in both columns, conditioning on B tells you nothing new about A. That flat line is a first look at independence, which we define precisely in C5–C6.

Quick check: Using the study-habits table from the Prerequisites section (36 students both study in a group and passed; 45 total group-studiers; 80 students total), which value is the denominator of ?

C3 — General Multiplication Rule

Rearranging the conditional probability formula gives a powerful tool for computing joint probabilities:

General Multiplication Rule

If , then

Symmetrically, if , then

Each form works for dependent or independent events when its conditioning event has positive probability. Use the form whose stated conditional probability is defined.

Derivation: Start from . Multiply both sides by :

The same rearrangement starting from gives the symmetric form.

Mini-example: A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Find .

. Given the first is red, (only 7 remain, 3 blue).

The simplified multiplication rule applies only to independent events. Do not use it unless you have verified independence first. For dependent events (like drawing without replacement), the General Multiplication Rule with the correct conditional probability is required.

C4 — Tree Diagrams for Multi-Stage Experiments

A tree diagram is a visual tool for applying the Multiplication Rule across sequential stages of an experiment.

Tree Diagram Rules

A probability tree represents a multi-stage experiment with one branch for each possible next outcome:

The probabilities on branches leaving any single node sum to 1.
The joint probability of a complete path is the product of the branch probabilities along that path.
The total probability of an event reached by multiple paths is the sum of the joint probabilities of those paths.

Labeling convention (use this in every problem):

Stage 1 branches: labeled with unconditional probabilities — and
Stage 2 branches: labeled with conditional probabilities — , , etc.
Path ends (not on branches): labeled with joint probabilities (the product of the branches along that path)

Figure 4: Stage 1 branches carry unconditional probabilities. Stage 2 branches carry conditional probabilities. Joint probabilities appear at the end of each path.

Try it yourself — complete the tree:

Enter your own branch probabilities, or click “Load Example” to pre-fill the factory scenario above. The path products compute automatically; each pair of branches from one node must sum to 1. To apply Rule 3, tick the leaves that make up an event — their path probabilities are added for you (the example pre-tags the two “Defective” paths, giving ).

Enter branch probabilities (0–1). Each pair from one node must sum to 1. Path probabilities = product of the two branches along that path.

Fill all inputs to verify the total sum.

Rule 3: tick the leaves that make up an event to add their path probabilities.

Figure 5: A two-stage probability tree. Complete each complementary branch pair, multiply along a path to find its joint probability, then tag and add the leaves that make up the event of interest.

Full example — factory with two assembly lines (tabular form):

A factory has two assembly lines. Line 1 produces 60% of output with a 2% defect rate. Line 2 produces 40% with a 5% defect rate.

Stage 1 (Line)	P(Line)	Stage 2 (Defect)	P(Defect \| Line)	Joint Probability
Line 1	0.60	Defective	0.02	0.60 × 0.02 = 0.012
Line 1	0.60	Good	0.98	0.60 × 0.98 = 0.588
Line 2	0.40	Defective	0.05	0.40 × 0.05 = 0.020
Line 2	0.40	Good	0.95	0.40 × 0.95 = 0.380

(sum of defective paths)

Check: all four joint probabilities sum to 1.000 ✓

Quick check: Which of these is best solved by drawing a full tree diagram, rather than applying the General Multiplication Rule once?

C5 — Independent Events: Definition

Independent Events

Events and are independent if:

When , this is equivalent to : knowing that occurred does not change the probability of . When , it is also equivalent to . Independence is a symmetric property — if is independent of , then is independent of .

Intuition: The marble-bag example in C1 showed that conditioning changed the probability (0.40 → 0.57). That means the events were dependent. For independent events, the conditional probability equals the unconditional probability — conditioning gives no new information.

Mini-example: Suppose and . If these are independent, then — knowing someone is left-handed tells us nothing about their coffee preference.

C6 — Independence Test and the Simplified Multiplication Rule

Independence Test and Simplified Rule

To test independence: compute and compare to .

If : events are independent
If : events are dependent

Simplified Multiplication Rule (for independent events only):

Equivalent check: When , compare to . Both methods give the same conclusion — use whichever is easier with the given information.

Mini-example: In a group of 200 people: 80 own a car (P = 0.40), 70 have a gym membership (P = 0.35), and 28 have both (P = 0.14).

Test: .

Conclusion: the events are independent.

Confirm: ✓

See the test visually: Drag P(A∩B) until the actual bar matches the expected bar — that is independence. Pull it higher or lower to see positive/negative dependence.

P(A) = 0.45

P(B) = 0.40

P(A∩B) = 0.3000

Expected under independence: P(A)·P(B) = 0.1800

P(A∩B) is bounded by the marginals: it can't exceed min(P(A), P(B)) or fall below P(A)+P(B)−1, so moving P(A) or P(B) may nudge this slider.

P(A∩B) = 0.3000 > P(A)·P(B) = 0.1800 → DEPENDENT (positively associated)

Figure 6: Two equivalent tests for independence. Compare the observed intersection with P(A)·P(B), and compare P(A | B) with P(A); independence requires the paired values to match.

C7 — Independence vs. Mutual Exclusivity

In L-PR-1, we noted that mutually exclusive events and independent events are not the same thing. Now we have the tools to be precise about why.

Property	Mutually Exclusive (ME)	Independent
Definition	— cannot co-occur
Joint probability	Always 0	(when independent)
Conditional probability	(if )	(if )
Relationship	ME events with , are always dependent	Independent events with , can never be ME

Proof that ME implies dependence: If and are mutually exclusive with and :

Since , the events are dependent. Knowing occurred makes impossible — the most extreme form of dependence.

“These events can’t happen at the same time, so they must be independent.” This is backwards. Mutually exclusive events (with positive probability) are the most extreme form of dependence — knowing one occurred makes the other impossible. Independence requires that gives no information about , which is the opposite of ME.

Both panels respond to “Condition on B” simultaneously — see what each type of relationship looks like when the sample space is restricted to :

P(A) = 0.40

P(B) = 0.35

Expected P(A∩B) if independent = P(A)·P(B) = 0.1400

The controls keep P(A) + P(B) ≤ 1 so the mutually exclusive panel always represents a possible probability model.

Mutually Exclusive → always dependent (knowing B ⇒ A is impossible; P(A|B) = 0 ≠ P(A))

Independent → never mutually exclusive (when P(A),P(B) > 0) (knowing B tells you nothing about A; P(A|B) = P(A))

Figure 7: Mutually exclusive events with positive probability cannot be independent. Conditioning on B eliminates A in the disjoint panel, but preserves A's original proportion inside B in the independent panel.

Quick check: , , . How would you classify this pair?

C8 — Asymmetry of Conditioning

In general, . The direction of conditioning carries meaning.

Medical testing example:

— sensitivity: 95% of people with the disease test positive
— positive predictive value: depends on prevalence

If the disease affects only 1% of the population and the test has both 95% sensitivity and 95% specificity, a person with a positive test has only about a 16.1% probability of actually having the disease (computed via a tree diagram — see Boss Fight Path A). The 95% sensitivity is real, but it answers a different question than the one a patient cares about.

The Prosecutor’s Fallacy: Reversing the conditioning direction — treating as if it were — is called the Prosecutor’s Fallacy. In a hypothetical courtroom example, an expert might state that “the probability of this DNA match given innocence is 0.0001,” while a jury could misinterpret that as “the probability of innocence given this match is 0.0001.” That reversal could contribute to a wrongful conviction because the two quantities answer very different questions.

“The test has 90% sensitivity, so if I test positive, I have a 90% chance of having the condition.” This confuses with . The actual probability of disease given a positive test depends on prevalence and specificity; it cannot be read directly from sensitivity. For rare conditions, the two values can differ dramatically.

See the numbers come alive: The grid below shows 400 people. Orange dots have the disease; blue dots are false alarms. Drag the prevalence slider down toward 1% — watch the orange (true positive) dots become vastly outnumbered by blue (false positive) ones, even as test accuracy stays fixed. Then click “Condition on Positive Test” to see only who tests positive.

True Positive — disease & test+ (ringed)

False Negative — disease & test−

False Positive — no disease & test+

True Negative — no disease & test−

Prevalence P(disease) = 1.0%

Sensitivity P(+|disease) = 90%

Specificity P(−|no disease) = 95%

P(disease) = 1.0%

P(+ | disease) = 90%

P(+ | no disease) = 5.0%

P(+) — Law of Total Prob = 5.85%

Expected in 400: 3.6 TP + 19.8 FP = 23.4 expected positive tests

P(disease | positive) = 15.4%

P(positive | disease) = 90% but P(disease | positive) = 15.4% — a difference of 75 pp

Figure 8: Base rates determine what a positive test means. Even a sensitive test can produce many more false positives than true positives when the condition is rare, so P(disease | positive) can be far below P(positive | disease). Dots are rounded for display; the formula panel preserves the exact expected-count calculation.

Quick check: A doctor says: “This test has 95% sensitivity — .” A patient who tests positive asks: “So I have a 95% chance of having the disease?” What does the doctor’s 95% actually tell us?

Section 4: Worked Examples

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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.

Example 1 — Fully Worked: Conditional Probability from a Two-Way Table

A company surveyed 200 employees on remote work preference and tenure length. The results:

	Prefers remote	Prefers in-office	Total
Tenure > 5 years	54	36	90
Tenure ≤ 5 years	66	44	110
Total	120	80	200

Find and .

I notice the first question asks “given tenure > 5 yr” — that tells me the conditioning event. The denominator will be the row total for long-tenure employees.

Part 1:

Numerator: cell count for (remote AND tenure > 5) = 54
Denominator: row total for tenure > 5 years = 90

Part 2:

Numerator: same cell — 54 (intersection is symmetric)
Denominator: column total for remote = 120

I pause to check: are these the same? . They share the same numerator but different denominators. The direction of conditioning changed the denominator — and therefore the answer. This is C8 in action.

Example 2 — Predicting Independence

A city surveyed 300 residents: 180 have an active gym membership, 120 report eating healthily most days, and 72 have both.

Before computing, predict whether gym membership and healthy eating are independent or dependent. Give one reason for your prediction.

Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.

Show Solution

Test using multiplication:

Since , the events are independent in this sample.

Confirm using conditional probability: ✓

The independence is confirmed by both tests. Note that this result does not mean gym membership and diet are causally unrelated — independence in a sample is a statistical finding, not a causal claim.

Example 3 — Tree Diagram with Less Support

A factory has two assembly lines. Line 1 produces 60% of parts with a 2% defect rate; Line 2 produces 40% with a 5% defect rate.

Before calculating, predict whether a defective part is more likely to have come from Line 1 or Line 2. Explain how the production shares and defect rates influence your prediction.

(a) Find . (b) Given a part is defective, find .

Show Solution

Tree diagram paths and joint probabilities:

Path	Joint Probability
Line 1 → Defective
Line 1 → Good
Line 2 → Defective
Line 2 → Good

(a)

(b)

Check your prediction: Even though Line 1 produces 60% of all parts, it accounts for only 37.5% of defective parts. A defective part is therefore more likely to have come from Line 2, whose higher defect rate outweighs its smaller production share.

Example 4 — Find the Error

A student reads: “Among people who exercised regularly, 80% also reported good sleep.” The student writes:

Student’s work:

“The problem states that 80% of regular exercisers have good sleep, so:"

"I can now use this to find the probability that a person who has good sleep also exercises.”

Before correcting the student’s work, predict whether the stated 0.80 alone is enough to compute the reversed conditional. Then identify the exact conditioning-direction error, rewrite the 0.80 statement correctly, and state what additional information would be needed.

Show Solution

The statement “80% of exercisers reported good sleep” translates to . The student has reversed the conditioning direction, writing instead.

These are different quantities:

— of exercisers, 80% sleep well
— of good sleepers, what fraction exercises? Requires additional data (base rates for exercise and sleep in the population)

Check your prediction: The stated 0.80 is not enough to compute the reversed conditional. The error is an instance of the Prosecutor’s Fallacy (C8): treating . The 0.80 cannot be assigned to the reversed conditional without knowing and .

Section 5: Guided Practice

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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 support now fades backward: the first two problems require the step schedule, the next two ask for a direct computation, and the final problem lets you choose a solo attempt or guided steps.

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

Identify the denominator of , then compute it.

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

Work through the full independence procedure one committed step at a time: the joint probability, the multiplication test, the conditional-probability confirmation, and a compound claim about mutual exclusivity versus independence.

Problem 3 — Applying the General Multiplication Rule (C3)

Given and , apply the General Multiplication Rule to find .

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

A two-stage experiment is described. Find the joint probability of a specified path by multiplying along the branches.

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

Return to the independence procedure from Problem 2 with less support. Choose the solo route to compute the diagnostic conditional directly, or ask for the two-step route to test independence before computing the conditional.

Section 6: Independent Practice

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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)

Compute , then compute from the same table — the same numerator, two different denominators.

(c) Are and always equal for two events and ?

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

A three-branch experiment with different conditional probabilities at stage 2. Find a joint probability and the total probability of a stage-2 outcome.

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

Given two probabilities and their intersection, test independence — then compute the conditional probability.

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

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

Work through the full loyalty-card/repeat-purchase dataset one committed step at a time: complete the table, compute both conditionals, test independence, then compute the reversed conditional and compare directions.

(e) A marketing manager says: “Since 50% of loyalty card holders made a repeat purchase, our loyalty program is working — half our loyal customers come back.” A skeptical analyst replies: “That number doesn’t tell us if the loyalty card caused the repeat purchase, and it doesn’t account for base rates.” Who is correct?

Section 7: Mastery Check

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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.

No hints. No scaffolds. Model answers are behind the disclosure triangle.

Question 1 — Feynman Test

Explain concisely to a classmate who missed today’s lesson why the probability of having a disease given a positive test result is not the same as the probability of testing positive given you have the disease. Use the words “conditioning direction,” make the different denominators explicit, give a concrete example where the two values are at least 0.50 apart, and finish with a clear conclusion.

Compare with a model answer

The conditioning direction specifies which event we already know happened. answers “given I got a positive test, what’s the chance I’m actually sick?” while answers “given I’m sick, what’s the chance the test catches it?” These are two different questions with two different denominators — one restricts the universe to positive-test people, the other to sick people.

Concrete example: A disease affects 2% of the population. The test has 90% sensitivity () and 95% specificity ().

Using a tree diagram on 10,000 people:

200 have the disease; 180 test positive (90% of 200)
9,800 don’t have it; 490 still test positive (5% false positive rate)
Total positive tests: 670

The two values: vs. — a difference of 0.63. Reversing the conditioning direction without accounting for base rates leads to dramatic overestimation of disease probability after a positive test.

Question 2 — Two-Stage Quality Control

A manufacturing plant uses a two-stage quality check. Stage 1 inspection catches 85% of defective items (). Items that pass stage 1 go to stage 2; stage 2 catches 70% of the defectives that reached it (). Suppose 4% of all items are defective.

(a) Which tool is most appropriate for organizing this problem?

Show Solution

Tree diagram for a defective item:

Stage 1	P	Stage 2	P (given passed S1)	Joint
Caught	0.85	(removed)	—	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

A tree diagram organizes exactly this kind of multi-stage, sequential structure — each branch carries a conditional probability given the branches above it.

(b) What is ?

(c) Are “caught at stage 1” and “caught at stage 2” independent events for a defective item?

Show Solution

The stages are dependent because stage 2 only sees items that passed stage 1. is a conditional probability on a restricted population — it cannot equal the unconditional because stage 2 never sees items caught at stage 1.

Question 3 — Error Analysis

A student’s work:

A study found that and . The student needs .

”Since there’s no reason these would be related, I’ll apply the multiplication rule: .”

The actual data shows .

Which statement best identifies the error?

Show Analysis

The simplified rule is valid only for independent events. The student reasoned from context (“no reason these would be related”) rather than running the test.

Correct procedure:

Compute
Compare to the actual
Since , the events are dependent

The correct joint probability is 0.20 — from the data, not from the formula. The formula cannot substitute for actual data; it can only confirm independence after the fact.

What makes this valid: is a consequence of independence — a fact you verify, not an assumption you make from context.

Self-Assessment

How confident are you with this material?

Still confusedReady for the Boss Fight

Section 8: Boss Fight

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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.

Choose your path. Both require integrating all eight concepts from this lesson.

🔬 Path A — The Medical Analyst

A clinical lab tests patients for a rare condition. Sensitivity and specificity are known. Compute the probability that someone with a positive test actually has the disease — and explain why the answer surprises.

📊 Path B — The Policy Analyst

A company uses an independence assumption in a decision model. Given a 2×2 frequency table, test the assumption mathematically, quantify any mismatch, and explain its practical consequence without claiming causation.

Path A — The Medical Analyst

The single workspace below carries four integrated tasks: organize both positive-test paths, reverse the conditioning direction for a positive result, analyze a negative result, and explain how the base rate changes the interpretation. Start solo if you can; choose guided steps if you need them. Either way, your first answer counts. ”🔄 Try a new case” gives you a fresh condition, prevalence, and test accuracy, not the same numbers again.

Reflection:

Pause and commit. On paper or in your notes, identify which part of your reasoning was hardest: incorporating the base rate, keeping the conditioning direction straight, or explaining the surprising result. State one check you will use the next time you interpret a test result, then compare with the model reflection.

Model Reflection

The hardest part for me was keeping the conditioning direction attached to the correct denominator. Sensitivity restricts attention to people who have the disease; positive predictive value restricts attention to everyone who tested positive. The base rate matters because, when the condition is rare, a modest false-positive rate can create many false positives outside the small diseased group. Next time I will write the target conditional in words first, build the relevant joint paths, and check that my denominator represents the group named after “given.”

Path B — The Policy Analyst

The single workspace below carries four integrated tasks: reconstruct the company’s predicted joint probability, compute the observed joint probability, test and quantify the mismatch, and advise the decision-maker about its practical consequence. Start solo if you can; choose guided steps if you need them. Either way, your first answer counts. ”🔄 Try a new case” swaps the entire scenario for a fresh one, not just new numbers.

Reflection:

Pause and commit. On paper or in your notes, identify which piece of evidence most changed your judgment: the predicted product, the observed joint probability, the error factor, or the practical consequence. Then name one additional check you would request before recommending a policy change and compare with the model reflection.

Model Reflection

The observed joint probability changed my judgment because it did not match the product predicted by independence. The error factor then made the practical size of that mismatch clear. I can conclude that the events are dependent and that an independence-based estimate is inappropriate, but dependence alone does not identify a cause. Before recommending a policy change, I would verify the table’s definitions and data quality, then look for relevant group differences or selection mechanisms that could explain the association.

Section 9: Challenge Problems

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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.

Ready for more? These go beyond the lesson objectives.

Challenge 1 — Algebraic Derivation

Assume , and start from the conditional probability formula .

(a) Rearrange the formula algebraically to derive the General Multiplication Rule. Show each step.

(b) If and are independent, use the conditional equivalent (valid here because ) in the General Multiplication Rule. What do you get?

(c) In a tree diagram, if stage 1 and stage 2 are independent experiments and the stage-1 branches under discussion have positive probability, which statement gives the defining property of independence for the stage-2 branch probabilities?

Challenge 2 — Redundant Inspection Systems

A quality-control line uses several independent inspectors; each one misses an existing defect with some probability . (A “miss” means the inspector fails to catch an existing defect.) Each problem below states its own value of — read it from the problem, not from memory of a previous part.

(a) Compute .

(b) Compute .

(c) How many inspectors are needed to reduce below 0.001?

Challenge 3 — Conditional Probability Across Three Groups

A three-group conditional probability problem. Compute conditional probabilities within each group and observe how they vary.

Section 10: Solutions Reference

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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.

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The solutions page covers every current generated, stepped, find-the-error, mastery, Boss Fight, and challenge item. For parameterized generators, it provides representative worked solution methods that apply across generated instances.

Quick-Reference Formulas — L-PR-2

Formula	Purpose	When to use
	Conditional probability	Any time you see “given” — requires
	General Multiplication Rule	Find joint probability from conditional — requires ; works for dependent or independent events
	Simplified Multiplication Rule	Independent events only — verify independence first
	Conditional form of independence	Equivalent test when

Key distinctions to remember:

vs. : Same numerator, different denominators. Not equal in general. Direction matters.
Mutually exclusive vs. independent: ME events (with positive probability) are always dependent. They are opposite concepts, not related ones.
Simplified vs. General Multiplication Rule: The simplified rule is a special case of the general rule, valid only when independence is confirmed.
from a table: Numerator = intersection cell; denominator = marginal total for (row or column total for the conditioning event, not the grand total).