L-PR-1: Basic Probability Concepts

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.

Imagine a hypothetical roulette player watching the wheel land on black nine times in a row. On the next spin, the player puts a large bet on red because red now feels “due.” Under the standard model of independent spins, however, the streak has not changed the chance of red on the next spin.

This mistake is called the gambler’s fallacy — treating a streak of independent outcomes as if it forces the next outcome to compensate.

The mathematics that exposes the mistake — and replaces intuition with a checkable argument — is probability. This lesson is where that reasoning begins.

In L-DS-1, you learned to read counts and relative frequencies from tables. Here, those same counts become the raw material for sample spaces, events, and probability rules. Understanding exactly why the roulette player was wrong also requires independence — the formal idea that one spin does not change the next spin’s probability. That definition arrives in L-PR-2; today you build the language it depends on.

After this lesson, you will be able to:

Identify the sample space of an experiment and express events as subsets of it.
Compute classical and empirical probabilities from equally likely outcomes and observed frequencies.
Apply the complement rule and the General Addition Rule to probabilities of compound events.
Identify mutually exclusive events using Venn diagrams and distinguish them from independent events.

This is Lesson L-PR-1. Everything that follows in this module — conditional probability, random variables, distributions — is built on the concepts in this lesson.

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-DS-1

Reading two-way frequency tables (L-DS-1, Sampling): L-PR-1 uses two-way tables as its primary data format. You need to read cell counts, row totals, and column totals accurately — today we convert those counts to probabilities.
Relative frequency as proportion (L-DS-1, Data Vocabulary): You already know that a relative frequency is a count divided by the total. Empirical probability is exactly this ratio applied to event outcomes.
Qualitative variables (L-DS-1, Variable Types): Most probability examples involve categorical outcomes (heads/tails, pass/fail, product A/B). Being fluent with categorical data makes the event language natural.

Retrieval Checkpoint

A school surveyed 120 students about their study method and exam result. The two-way table below shows the results.

	Passed	Failed	Total
Studied in groups	48	12	60
Studied alone	42	18	60
Total	90	30	120

Also confirm you can recall:

I can identify the row total and column total for any cell in a two-way table. I know that a relative frequency is always between 0 and 1. I can convert a count to a proportion by dividing by the appropriate total.

Success Factor:

From description to prediction. In L-DS-1, relative frequencies described data that had already been collected — they summarized what had happened. In L-PR-1, we use the same count-over-total structure to describe the chance of an event. The arithmetic is familiar; the interpretation is new.

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 — 10 Concepts

C1–C4: What probability means — four ways to assign numbers to uncertain outcomes.
C5–C6: The rules every probability must obey, and the most useful shortcut: the complement.
C7–C9: Combining two events with “or” and “and,” culminating in the General Addition Rule.
C10: A critical distinction that trips up students all semester: mutually exclusive ≠ independent.

C1 — Sample Space and Events

Every probability problem begins with an experiment — any process whose outcome is uncertain. We write for the sample space: the set of all possible outcomes. We write to say that event is a subset of — a collection of outcomes that interest us.

Sample Space and Event

The sample space is the set of all possible outcomes of an experiment.

An event is any subset of : .

Event occurs when the actual outcome belongs to .

Example. Roll a fair six-sided die. The sample space is . Let = “roll an even number” . If the die shows 4, event occurs.

C2 — Classical Probability

When every outcome in is equally likely, probability is a simple counting ratio.

We write for the number of outcomes in event .

Classical Probability

For an experiment with equally likely outcomes:

Example. With the die above: , , so .

Classical probability requires equally likely outcomes. If you flip a bent coin, the sample space is still , but — the formula gives the wrong answer for non-uniform spaces. A particularly tempting error: “It will rain tomorrow or it won’t — so there’s a 50% chance of rain.” That reasoning applies to a space where outcomes are not equally likely. Rain tomorrow is an empirical or subjective probability, not a classical one.

C3 — Empirical Probability

When we cannot assume equally likely outcomes, we estimate probability from observed data.

Empirical (Relative Frequency) Probability

After trials, the empirical probability of event is:

where is the number of trials in which occurred. If the trials are independent and repeated under the same conditions with a fixed event probability, this estimate converges toward that probability as grows (Law of Large Numbers — conceptual).

Example. A factory inspects 400 items and finds 20 defective. The empirical probability of a defective item is .

Try it yourself. The top chart illustrates how the running proportion tends toward 0.50 as grows. The bottom chart shows the gap between the number of heads and tails. A particular gap can rise, fall, or return to zero; its typical size is on the order of , so the dashed curve is a scale guide rather than a forecast of the path. That contrast is the whole point: a run of tails (watch the recent-flips strip) never makes the line “correct” upward. And no matter how long the current streak gets, the predictor below it keeps reading — the coin has no memory. Convergence comes from averaging, not from the counts catching up.

Recent flips:

Click a button to start flipping.

Figure 1. The running proportion of heads settles toward 0.50 even while the absolute heads–tails gap keeps wandering. Recent streaks do not change the next-flip probability: convergence comes from averaging, not from the counts “catching up.”

C4 — Subjective Probability

A subjective probability expresses a person’s degree of belief, informed by the evidence and judgment available to that person. It is especially useful for one-time or poorly repeatable events, but it can also incorporate empirical evidence.

Enrichment — beyond the assessed standards. Subjective probability — and classifying a probability as classical vs. empirical vs. subjective — is here for context: it sharpens how you read probability claims in the world. The graded standard (S-PR-1) assesses computing classical and empirical probabilities and applying the rules, not naming which type a number is. Read this to think clearly; it won’t appear on a standards assessment.

Subjective Probability

A subjective probability is a numerical degree of belief that an event will occur, assigned by a person based on their knowledge and judgment. It must still satisfy and (C5).

Example. A venture capitalist says: “I estimate a 30% probability this start-up will reach profitability within two years.” No equally likely counting rule determines this number. The estimate may combine base rates, evidence about this company, assumptions, and informed judgment.

Every probability assignment — whether classical, empirical, or subjective — must obey the same axioms (C5). A subjective probability that “feels right” but violates is internally inconsistent and must be revised.

For each scenario below, decide which approach best describes how the probability is assigned before checking — then pick a card to see if you were right. Real judgments can combine approaches; choose the primary one in each example.

Pick the primary probability approach for each scenario below.

Choose the approach that best describes how the probability is assigned:

Figure 2. A predict-then-check comparison of classical, empirical, and subjective probability. The classifier is enrichment: L-PR-1 assesses computation with classical and empirical probabilities, not naming the three types.

C5 — Probability Axioms

All valid probabilities, regardless of how they are computed, satisfy three foundational axioms.

Probability Rules

For events in a sample space :

Non-negativity: for every event .
Normalization: — the complete sample space has probability 1.
Additivity: If , then .

Two immediate consequences: , and every event satisfies .

For the curious: the formal Kolmogorov third axiom applies to any countable collection of pairwise disjoint events. The two-event version above is all this lesson needs.

Validity check. If pass and fail are the only two outcomes, they are disjoint and together equal . A claim that and gives , contradicting normalization and additivity. Valid complementary events must sum to exactly 1.

Every valid probability lives in the interval — but is not a probability. Drag and below: their raw sum (the upper needle) can shoot past 1 into the red zone, and that breaks no axiom — it simply means the sum has counted the overlap twice. Now drag : subtracting it once pulls the lower needle, , back into the valid interval, where a real probability always lives. That subtraction is exactly the General Addition Rule (C8).

not a probability

0 Impossible 0.5 Fair coin 1 Certain

P(A) 0.60

P(B) 0.50

P(A∩B) 0.30

Figure 3. The raw sum

may extend past 1 because it double-counts overlap; subtracting

moves the union

into the valid probability interval.

C6 — Complement Rule

The complement of event consists of all outcomes in not in . Because and together cover all of and never overlap, their probabilities must sum to 1.

Complement Rule

Strategy: When computing directly is hard, compute and subtract from 1. This is especially powerful for “at least one” problems.

Example. , so .

Checkpoint — C5 and C6

Question A. Apply the complement rule to find the missing probability.

Question B. A completed game has exactly two outcomes: win or lose. A classmate says: ” and too — both outcomes are possible, so both can have the same probability.” What is the problem with this claim?

C7 — Compound Events: Union and Intersection

Two events can be combined using set operations. Understanding these operations precisely prevents the most common errors with the Addition Rule.

Union and Intersection

— the union of and : the event that occurs, or occurs, or both. (At least one.)

— the intersection of and : the event that both and occur simultaneously.

Example. Roll a die. Let (even) and (greater than 3).

— even or greater than 3 (or both).
— even and greater than 3 simultaneously.

Venn diagram: draw two overlapping circles inside rectangle . The left-only region is only; the right-only region is only; the overlap is ; outside both circles is neither.

Click any cell in the table below to see how each table entry maps onto the four Venn regions. This connection bridges the L-DS-1 frequency table format you already know with the event language used in all probability calculations. Use “Show as probabilities” to flip every count to its proportion (count ÷ 120) — the same diagram, now reading where it read .

Scenario: A school classified 120 students by their primary study method and whether they passed an exam. Each student has exactly one primary method. Select one student at random. A means “group was the primary study method”; B means “passed the exam.”

Survey of 120 students — click any cell

	Passed B	Failed B′	Total
Group A	48	12	60
Individual A′	42	18	60
Total	90	30	120

Venn diagram (n = 120)

Click any highlighted cell or total to see the corresponding Venn region.

Figure 4. For one randomly selected student,

means “studied in groups” and

means “passed the exam.” Each two-way-table cell maps to one Venn region. Switching from counts to probabilities changes the units, not the regions or their relationships.

C8 — General Addition Rule

When you add , every outcome in gets counted twice — once as part of and once as part of . The General Addition Rule corrects for this double-counting.

General Addition Rule

For any two events and :

The intersection is subtracted once to undo the double-counting of outcomes that belong to both events.

Example. In a class of 40 students: 18 play a sport, 14 are in a club, and 6 do both.

Without subtracting: — wrong, because the 6 students in both groups were counted twice.

Never add the intersection — always subtract it. A frequent error is . Adding the intersection makes the double-counting worse, not better.

You cannot compute by multiplying unless you know and are independent. The formula requires independence — a condition formally defined in L-PR-2. In L-PR-1, always read directly from the data or the problem statement.

Try dragging the sliders below. Each dot is one of 100 students ( = plays a sport, = is in a club) — use “Show as sets” to fade from that concrete picture to the abstract , notation once the idea is clear. Notice that is always less than when the events overlap — because the intersection gets counted twice without the correction. Click “Make ME” to see what changes when the events cannot both happen.

Each dot is 1 of 100 students. A = plays a sport, B = is in a club.

|A| 30

|B| 20

|A ∩ B| 10

Figure 5. One hundred dots make the General Addition Rule visible: the intersection belongs to both events, so adding

and

counts those outcomes twice and the rule subtracts the overlap once.

C9 — Mutually Exclusive Events

Two events are mutually exclusive (ME) if they cannot both occur in a single trial — their intersection is empty.

Mutually Exclusive Events

Events and are mutually exclusive if their intersection is empty:

Consequently, .

In the finite sample spaces used in this lesson, where every possible outcome has positive probability, a zero-probability intersection also confirms that the intersection is empty.

For mutually exclusive events, the General Addition Rule simplifies to:

Example. Roll a die. Let = “roll a 1” and = “roll a 6.” One trial cannot show both, so . Therefore .

Always verify mutual exclusivity before using the simplified Addition Rule. Mutual exclusivity is a mathematical condition — — not a semantic one. Two events about “different things” can still share outcomes. Example: “owns a car” and “uses transit” seem unrelated, but one commuter can do both. If you apply to overlapping events, the result may exceed 1, violating the probability bounds.

C10 — Mutually Exclusive vs. Independent (Preview)

These two terms describe completely different relationships. Mixing them up is one of the most persistent errors in this course.

Mutually exclusive is about overlap in the sample space: — the events cannot co-occur.

Independent is about informational influence: knowing one event occurred does not change the probability of the other. Formally, . When , this is equivalent to . Independence is defined and tested in L-PR-2.

ME vs. Independence — Critical Distinction

Mutually exclusive: — the events cannot happen at the same time.

Independent (preview): . When , the equivalent conditional statement is .

Two events with positive probability that are mutually exclusive are always dependent. If occurred, definitely did not — so knowing changes the probability of from to 0. That is the opposite of independence.

Notation boundary. We say and are independent when knowing occurred does not change the probability of . This is a precise mathematical condition — not the same as “they seem unrelated.” We will define and test it formally in L-PR-2. For now: do not use “mutually exclusive” and “independent” as synonyms.

“These events are mutually exclusive, so they must be independent — they can’t affect each other.” This is backwards. If and are mutually exclusive with and , knowing occurred tells you with certainty that did not. The events are necessarily dependent.

Click “A occurred →” in each panel to see how the two concepts differ — one rules out entirely, the other leaves completely unaffected.

Mutually Exclusive

Die roll — A = {1, 2}, B = {4, 5, 6}

Independent

Two coins — A = Flip 1 = H, B = Flip 2 = H

Figure 6. When mutually exclusive events have positive probability, observing

rules out

; for independent events, observing

leaves the probability of

unchanged.

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: Classical Probability with a Standard Deck

Problem. A card is drawn at random from a standard 52-card deck. Let = “draw a heart” and = “draw a face card (J, Q, K of any suit).” Compute , , , , and .

Metacognitive narration.

I notice that the deck is fair, so all 52 cards are equally likely. That tells me to use classical probability: count the outcomes in each event and divide by 52.

I need to count the two events and their overlap before choosing an addition rule.

, so
, so
, so

I notice that and overlap at J♥, Q♥, and K♥. Because the intersection is not empty, I choose the General Addition Rule rather than the mutually exclusive shortcut.

For the complement, I subtract from 1 because every card is either a face card or a non-face card:

The result is the probability of drawing a non-face card. The complement shortcut is faster than counting those 40 cards one at a time.

Example 2 — Empirical Probability from a Two-Way Table

Problem. A university surveyed 200 students about streaming subscriptions. Results:

	Has Netflix	No Netflix	Total
Has Spotify	60	36	96
No Spotify	70	34	104
Total	130	70	200

Let = “has Netflix” and = “has Spotify.”

From the table, find , , and . Then predict and explain how the overlap affects your answer.

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

Read each probability using the grand total of 200:

Apply the General Addition Rule:

Check: Adding only gives , which cannot be a probability. The 60 students with both subscriptions were counted once in the Netflix total and once in the Spotify total, so the intersection must be subtracted once.

Example 3 — Complement After a Union

Problem. In a group of 500 employees: , , . Before calculating, predict whether more or less than half of the employees have neither arrangement. Then find (a) and (b) .

Show Solution

Part (a) — Find the union.

Part (b) — Take its complement.

Check your prediction: , so less than half of the employees have neither arrangement.

Pattern to remember: , and . This two-step pattern appears constantly in probability problems.

Example 4 — Independent Application: What the Labels Do Not Tell You

Problem. Among 1,000 website sessions, 620 were on mobile devices, 410 included a purchase, and 280 were both mobile and included a purchase. For one randomly selected session:

Before calculating, predict whether adding directly would overcount, undercount, or give the exact union. Explain which group in the data determines your answer.

Find the probability that it was mobile or included a purchase.
Find the probability that it was neither mobile nor included a purchase.
Decide whether “mobile” and “purchase” are mutually exclusive. Justify from the data.

Show Solution

Let = “mobile” and = “purchase.”

Part 1 — Find the union.

Check your prediction: Adding directly would overcount the 280 sessions in the intersection, because each is included once in the mobile total and once in the purchase total.

Part 2 — Find neither.

Part 3 — Check mutual exclusivity.

Because , the events are not mutually exclusive. The 280 sessions in the intersection prove that both events can occur together.

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 fades across this section: the first problem forces a step-by-step schedule, then the remaining problems ask you to produce the result directly.

Problem 1 — Venn Diagram and Mutual Exclusivity (C7 + C8 + C9 + C10)

Work through the first two computations one step at a time:

Part (c). Are “plays sport” and “plays instrument” mutually exclusive? Justify with a probability.

Part (d). A classmate says: “Since playing sport and playing instrument are two completely different activities, they must be independent.” Choose the best response.

Problem 2 — Classical Probability and Complement (C1 + C2 + C6)

Problem 3 — Applying the Complement Rule (C5 + C6)

Problem 4 — General Addition Rule (C7 + C8)

The following problem gives you , , and in a realistic context. Apply the General Addition Rule to find . Note that in every variant — the events always overlap, so the simplified rule does not apply.

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 — Empirical Probability from a Two-Way Table (C1 + C3)

Problem 2 — Addition Rule from a Two-Way Table (C3 + C7 + C8)

Problem 3 — Complement Strategy: At-Least-One Problems (C5 + C6)

These problems involve multiple independent trials. The problem statement notes that trials are independent — the formal definition arrives in L-PR-2, but you can use it intuitively here: the outcome of one trial does not affect the others. Your task is to recognize why the complement strategy is more efficient than listing every case directly.

Problem 4 — Find the Error: Probability Reasoning (C8 + C9 + C10)

Problem 5 — Multi-Step Synthesis (C1 + C3 + C6 + C7 + C8 + C9 + C10)

A class of 120 students is categorized by two attributes: whether they submitted all assignments (yes: 80, no: 40) and whether they passed the final exam (yes: 90, no: 30). Among those who submitted all assignments, 75 passed the final.

Complete the table and the probability computations independently. The answer box checks the final “neither” probability, but you will need the earlier results to reach it.

Part (d). Are “submitting all assignments” and “passing the final” mutually exclusive? Justify.

Part (e). A classmate says: “Submitting assignments and passing are different things, so they must be independent.” Evaluate this claim.

Mixed Review — Retrieval from Earlier Lessons

These problems draw on concepts from earlier in the course. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.

Review Problem 1 — Range (L-DS-4)

Review Problem 2 — Empirical Rule (L-DS-5)

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. These questions test whether you can recall and apply what you have learned without support.

Question 1 — Feynman Test

Explain to a classmate who missed today’s lesson why we cannot always compute . Use a concrete example — numbers or a deck of cards — where this formula gives the wrong answer, show the correct calculation, name the double-counted overlap, and state the condition that makes the shorter rule valid.

Compare with a model answer

Imagine a class of 30 students: 12 play piano, 10 play guitar, and 4 play both. If you add and divide by 30, you get 22/30 — but those 4 who play both got counted twice: once in “piano” and once in “guitar.” The true count of students who play at least one instrument is , giving . The General Addition Rule fixes the double-counting by subtracting the intersection once. The simple formula is only valid when the events are mutually exclusive — when no outcome can belong to both at the same time.

Question 2 — Apply

Question 3 — Error Analysis

A student writes: ” and . Since these events seem unrelated, I will apply the simple Addition Rule: .”

Student’s conclusion: ” — this means it is very likely that at least one of the two events occurs.”

Identify the error in the student’s work.

Show Solution and Correction

Error type: Applied the simplified Addition Rule without verifying that .

Why the result is impossible: violates the probability bound . This is the clearest possible signal that the simplified rule was applied incorrectly.

Correction: Without knowing , we cannot compute exactly. We can establish bounds: and . The actual answer requires from the data.

Self-Assessment

How confident are you with the material in this lesson?

Still confused (0)Ready for the Boss Fight (100)

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:

📊 Path A — The Pollster

A citywide survey lands on your desk with a colleague’s confident verdict attached. Is the colleague right?

🎙️ Path B — The Referee

A commentator does probability math live on air and arrives at 115%. Something’s off — find it.

📊 Path A — The Pollster

A citywide survey of 500 residents crossed gym membership with healthy-eating habits. A colleague skims the numbers and announces to the room:

“Gym members and healthy eaters are basically separate crowds — one group is about exercise, the other about diet. They barely overlap.”

Here is the full table:

	Healthy Eating	Not Healthy Eating	Total
Gym Member	110	90	200
Not a Member	130	170	300
Total	240	260	500

Task 1. For one randomly selected resident, find , , and .

Task 2. Find and interpret the result for these 500 residents.

Task 3. Find the probability that a resident is neither a gym member nor a healthy eater. Verify the result against one table cell.

Task 4. Evaluate the colleague’s “separate crowds” claim. Are the events mutually exclusive? Use the formal condition and the table evidence.

Commit before checking. Complete all four tasks on paper, in a notes app, or in your course notes. Your response should contain the three probabilities from Task 1, the two computed results, and a one-sentence verdict on the colleague’s claim.

Show Solution

Task 1.

— this uses the intersection cell, not the row or column total.

Task 2. Interpretation: 66% of the 500 residents are gym members, or eat healthily, or both.

Task 3. . Verify: “Not a member AND not healthy eating” cell = 170/500 = 0.34. ✓

Task 4. Mutual exclusivity requires . But . The events are not mutually exclusive — 110 residents are both gym members and healthy eaters. The colleague’s semantic reasoning (“separate things”) does not substitute for the formal mathematical check.

Reflect after comparing. What would the table have to look like if gym membership and healthy eating were mutually exclusive?

Compare with a model reflection

The “gym member AND healthy eating” cell would have to be 0. The row and column totals could still be positive, but no resident could appear in both groups. Then , and the simplified Addition Rule would apply.

🎙️ Path B — The Referee

A sports commentator says on live television:

“The home team wins 60% of the time and scores first 55% of the time — so their chance of winning OR scoring first is 115%! They are basically unstoppable!”

Task 1. Before choosing a formula, identify the fastest mathematical sign that 115% cannot be the probability of a union.

Task 2. Identify the rule the commentator used and the condition that would have to be true for that shorter rule to apply.

Task 3. Name the missing probability. Then suppose the team wins and scores first in 40% of games; compute the correct union probability.

Task 4. Explain why “winning” and “scoring first” being different descriptions does not make them mutually exclusive.

Commit before checking. Complete all four tasks on paper, in a notes app, or in your course notes. Include the corrected probability and a sentence explaining exactly where the commentator double-counted.

Show Solution

Task 1. violates the probability bound: all probabilities must lie between 0 and 1. The 115% result is enough to prove that the calculation is wrong.

Task 2. The commentator applied the simplified Addition Rule: . This rule requires — that winning and scoring first cannot happen in the same game. That condition is false: both events can occur in the same game.

Task 3. The missing piece is — the probability of winning AND scoring first in the same game.

With : . The correct probability is 75%, not 115%.

Task 4. “Measuring different things” does not imply mutual exclusivity. A single game can result in both a win and a first score for the home team. Mutual exclusivity requires that both events literally cannot occur in the same trial — which is clearly not the case here. The correct bound is ; the actual value requires the intersection probability.

Reflect after comparing. What single question could you ask whenever someone adds two probabilities to decide whether an overlap must be subtracted?

Compare with a model reflection

Ask: “Can both events happen on the same trial?” If yes, the events are not mutually exclusive, so the intersection can be positive and must be accounted for with the General Addition Rule.

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 — Derive the General Addition Rule from Counting (C7 + C8 + C9)

The General Addition Rule is . Don’t take it on faith — build it from counting.

For a sample space of equally likely outcomes: start by counting the outcomes in in terms of , , and , and explain why one term must be subtracted rather than added. Then turn your count into probabilities by dividing through by . Finally, check what your formula becomes when and are mutually exclusive.

Work on paper or in your notes, then compare your derivation with the solution.

Show Solution

Count the union. Add and and every outcome that lives in both and has been counted twice — once inside each set. Subtract that overlap exactly once to fix the double-count:

Why subtract, not add? Adding would count those shared outcomes three times. And it is not — multiplication counts ordered pairs of sequential choices, not the outcomes of a single draw.

Divide by . Every count becomes a probability:

The mutually exclusive case. If and cannot both happen, no outcome is shared, so and the subtracted term vanishes: . The simplified rule is not a different rule — it is the general rule with a zero overlap.

Challenge 2 — Non-Uniform Probability Space (C1 + C2 + C5 + C8)

A biased die has . The remaining five faces (1, 2, 3, 4, 5) are equally likely.

Challenge 3 — A Hidden Simplification (C7 + C8)

Some unions are easier than they look. Read the setup carefully before you reach for a formula — how and relate is the whole puzzle.

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.

Complete, step-by-step solutions for all problems in Sections 5–9 are available on the solutions page.

View Full Solutions →

If you are stuck: Re-read the Core Concept in Section 3 that applies, then find the Worked Example in Section 4 that maps to it. The solutions page shows reasoning behind every step, not just the final answer.

Quick-Reference Formulas

Probability Axioms:

Immediate consequences:

Complement Rule:

General Addition Rule:

Mutually Exclusive Events (when ):

Mutually Exclusive vs. Independent:

ME: — overlap is impossible
Independent (L-PR-2): ; if , equivalently
ME events with positive probability are always dependent

Back to the roulette wheel. You now have the tools to understand the imagined player’s mistake. On a fair wheel, each spin is an independent event — knowing the last 9 results gives no information about the next spin. That is a precise mathematical condition (), formally defined in L-PR-2. The player’s error was treating nine consecutive blacks as evidence that red was “due,” as if the spins were dependent. With the probability rules from this lesson and the independence definition from L-PR-2, you will be able to diagnose that error precisely.