Conditional Probability

Section 1: Introduction

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

This is a conditional probability question — and the answer might surprise you. When COVID was rare in the population (say, 1% prevalence), a positive result on a 95%-accurate test meant the chance of actually having the disease was far lower than 95%. The mathematics behind this apparent paradox is the subject of today’s lesson.

In PR-1, you learned to compute — the probability that both events occur. That joint probability is a building block. In 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 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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What you need from PR-1

Reading joint and marginal frequencies from two-way tables (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 (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 (PR-1, Core Concepts C10): 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

What is ?

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

Section 3: Core Concepts

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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 PR-1 preview: why mutually exclusive events are never independent.
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 .

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 B” to collapse the non-B row. Watch the denominator move from the grand total to B’s row total — that shift is the entire difference between and .

	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 P(A|B)

If you divide by n = 80 you get P(A∩B) = 0.45, not P(A|B). Condition on B to restrict the universe.

The non-B row collapses — B becomes the new universe

Explore the formula interactively:

|A| = 30

|B| = 25

|A∩B| = 10

P(A) = 0.30 P(B) = 0.25 P(A∩B) = 0.10

P(A | B) = P(A∩B) / P(B) = 0.10 / 0.25 = 0.40

Drag the sliders to adjust the sizes of A, B, and their overlap. Then click “Condition on B” to restrict the sample space to B. Notice how changes when B is large vs. small — even for the same overlap. The independence indicator lights up green when conditioning on B tells you nothing new about A.

C3 — General Multiplication Rule

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

General Multiplication Rule

This works for any two events — dependent or independent. Use it whenever you know a conditional probability and need the corresponding joint probability.

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.

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)

Three rules for tree diagrams:

Probabilities on branches from any single node must sum to 1.
The joint probability of a complete path = product of its branch probabilities.
The total probability of an event across multiple paths = sum of the joint probabilities of all paths that lead to it.

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.

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

Sum: —

Outcome A

Outcome B

Sum: —

A ∩ X ?

A ∩ Y ?

B ∩ X ?

B ∩ Y ?

Fill all inputs to verify the total sum.

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 ✓

C5 — Independent Events: Definition

Independent Events

Events and are independent if:

Knowing that occurred does not change the probability of . Equivalently, . 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: 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) = 0.18

Expected under independence: P(A)·P(B) = 0.180

P(A∩B) = 0.18 = P(A)·P(B) = 0.180 → Events are INDEPENDENT

C7 — Independence vs. Mutual Exclusivity

In 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	— gives no info about
Joint probability	Always 0	(when independent)
Conditional probability	(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) = 0.140

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))

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, a person with a positive test may still have only a ~16% probability of actually having the disease (computed via 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. It has led to wrongful convictions when expert witnesses stated that “the probability of this DNA match given innocence is 0.0001” and the jury interpreted it as “the probability of innocence given this match is 0.0001.” These are very different quantities.

“The test is 90% accurate, 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 cannot be read directly from the accuracy rate. 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+

False Negative — disease & test−

False Positive — no disease & test+

True Negative — no disease & test−

Prevalence P(disease) = 2.0%

Sensitivity P(+|disease) = 90%

Specificity P(−|no disease) = 95%

P(disease) = 2.0%

P(+ | disease) = 90%

P(+ | no disease) = 5.0%

P(+) — Law of Total Prob = 6.70%

Of 400 people: 7 TP + 20 FP = 27 positive tests

P(disease | positive) = 26.2%

P(positive | disease) = 90% but P(disease | positive) = 26.2% — a difference of 64 pp

Section 4: Worked Examples

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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 — Prediction Checkpoint: Independence Test

A gym surveyed 300 members: 180 have a gym membership, 120 report eating healthily most days, and 72 have both.

Before computing: Based on context, do you think gym membership and healthy eating are independent or dependent? Write your prediction and reasoning, then check below.

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 — Details/Summary: Tree Diagram

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.

(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)

Even though Line 1 produces 60% of all parts, it accounts for only 37.5% of defective parts — because its defect rate (2%) is much lower than Line 2’s (5%).

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.”

Identify and correct the error:

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)

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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