Probability Fundamentals — SL 4.3 Probability fundamentals (sample space, events, Venn diagrams, tree diagrams) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Humans have been gambling for thousands of years, but for most of that time nobody had a systematic way to calculate their chances of winning. Dice carved from animal bones have been found in archaeological sites dating back over 5,000 years, yet the mathematical study of probability only began in the 17th century. The spark came from a simple question about a dice game, and the ideas that followed now underpin everything from weather forecasting to medical testing to machine learning.

1654 • The Problem of Points

French gambler Antoine Gombaud (the Chevalier de Méré) posed a question about splitting stakes in an interrupted dice game to mathematician Blaise Pascal. Pascal and Pierre de Fermat exchanged letters that laid the foundations of probability theory.

1713 • Ars Conjectandi

Jacob Bernoulli published Ars Conjectandi, which introduced the law of large numbers and established probability as a rigorous branch of mathematics.

1880 • Venn Diagrams Introduced

English logician John Venn popularized overlapping circle diagrams to visualize logical relationships between sets — a tool that became essential for representing events in probability.

1933 • Kolmogorov's Axioms

Russian mathematician Andrey Kolmogorov published a set of three axioms that put probability on a solid mathematical foundation. These axioms are still the basis of all modern probability theory.

The central question probability answers is straightforward: how likely is a particular outcome? To answer it, we need a clear language for describing all possible outcomes, the specific outcomes we care about, and the tools — like Venn diagrams and tree diagrams — that help us organize and calculate. That is exactly what IB SL 4.3 covers.

SECTION 2

Core Principles & Definitions

Before you can calculate any probability, you need to understand a handful of foundational ideas. These concepts form the vocabulary of the entire topic, and they appear in virtually every probability question you will encounter on the IB exam.

Experiment & Outcome

An experiment (or trial) is any process whose result is uncertain, like rolling a die. Each possible result is called an outcome.

Sample Space (U)

The sample space U is the set of all possible outcomes. For a standard die, U = {1, 2, 3, 4, 5, 6}. The IB often uses U to denote it.

Event (A, B, …)

An event is a subset of the sample space — one or more outcomes grouped together. For example, "rolling an even number" = {2, 4, 6}.

Complementary Event (A′)

The complement A′ contains every outcome in U that is not in A. So P(A′) = 1 − P(A).

Equally Likely Outcomes

When every outcome in U has the same chance of occurring, the probability of event A is simply P(A) = n(A) ÷ n(U), where n(·) counts elements in a set.

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KEY TAKEAWAY

Think of the sample space as a deck of cards laid face-up on a table. Every card is a possible outcome. An event is any group of cards you circle with a marker — maybe all the hearts, or all the face cards. The probability of your event is just the fraction of the table you circled. Everything else is the complement.

SECTION 3

Venn Diagrams — Visualizing Events

A Venn diagram uses overlapping circles inside a rectangle to show how events relate to each other and to the sample space. The rectangle represents U, and each circle represents an event. The region where circles overlap represents outcomes belonging to both events simultaneously — the intersection A ∩ B. The total area covered by both circles represents outcomes in at least one event — the union A ∪ B.

The cyan circle is event A, the violet circle is event B, and the pink overlap region is A ∩ B (the intersection). The area outside both circles but inside the rectangle represents outcomes in neither event.

The Venn diagram makes the addition rule visually obvious: if you add P(A) + P(B), you count the overlap region twice, so you must subtract it once. That gives P(A ∪ B) = P(A) + P(B) − P(A ∩ B). You can literally see this on the diagram — try shading A, then shading B, and notice the intersection gets double-shaded.

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IB Notation Note

The IB uses A′ for the complement of A (some textbooks write Ā or A^c). Make sure you are comfortable with A′ notation, as it appears throughout IB papers.

SECTION 4

Mathematical Framework

The equations below are the core formulas you need for SL 4.3. Each one translates a visual idea — from Venn diagrams or counting — into a precise calculation.

CLASSICAL PROBABILITY

P(A) = n(A) / n(U)

n(A) = number of outcomes in event A; n(U) = total number of equally likely outcomes in the sample space.

COMPLEMENT RULE

P(A′) = 1 − P(A)

The probability that A does not occur. Since all probabilities in U sum to 1, what's left over after A is its complement.

ADDITION RULE (UNION)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

P(A ∪ B) = probability of A or B (or both). Subtract the intersection to avoid double-counting the overlap.

MUTUALLY EXCLUSIVE EVENTS

If A ∩ B = ∅, then P(A ∪ B) = P(A) + P(B)

When two events cannot happen at the same time (their circles don't overlap in the Venn diagram), the intersection is empty and the formula simplifies.

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Probability Bounds

Every probability satisfies 0 ≤ P(A) ≤ 1. A probability of 0 means the event is impossible; a probability of 1 means it is certain. If you ever calculate a probability outside this range, you have made an error — go back and check your work.

SECTION 5

Tree Diagrams — Sequential Events

While Venn diagrams shine when you have two overlapping events, tree diagrams are the go-to tool when events happen in sequence — one after the other. Each branch represents a possible outcome at a given stage, and you write the probability of that outcome along the branch. To find the probability of a specific path from start to finish, you multiply along the branches. To find the probability of multiple paths that all satisfy the same condition, you add across the paths.

A tree diagram for drawing two marbles without replacement from a bag of 3 red and 2 blue. The first branch shows the initial draw; the second shows the conditional draw. Multiply along each path and verify that all paths sum to 1.

Notice how the second-stage probabilities change because the marble is not replaced. After drawing a red marble first, only 2 red and 2 blue remain, so the probability of red on the second draw becomes 2/4 rather than 3/5. This is what "without replacement" means in practice, and tree diagrams make it easy to track these changing probabilities stage by stage.

✓ Multiply along a path to get the probability of that specific sequence of outcomes.
✓ Add across paths when you want the probability of any one of several sequences (e.g., "at least one red").
✓ All paths must sum to 1 — this is a great self-check for your work.

SECTION 6

Worked Example — Venn Diagram Problem

In a class of 30 students, 18 study Biology (B) and 12 study Chemistry (C). 7 students study both. A student is selected at random. Find the probability that the student studies Biology or Chemistry (or both), and the probability that the student studies neither subject.

Venn Diagram with Two Events

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Step 1 — Identify the given values

n(U) = 30, n(B) = 18, n(C) = 12, n(B ∩ C) = 7.
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Step 2 — Fill in the Venn diagram regions

Only Biology (not Chemistry): 18 − 7 = 11. Only Chemistry (not Biology): 12 − 7 = 5. Both: 7. Neither: 30 − (11 + 7 + 5) = 30 − 23 = 7.

Regions: only B = 11, B ∩ C = 7, only C = 5, neither = 7
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Step 3 — Calculate P(B ∪ C) using the addition rule

P(B ∪ C) = P(B) + P(C) − P(B ∩ C) = 18/30 + 12/30 − 7/30 = 23/30.

P(B ∪ C) = 23/30 ≈ 0.767
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Step 4 — Calculate P(neither) using the complement rule

P(neither) = P((B ∪ C)′) = 1 − P(B ∪ C) = 1 − 23/30 = 7/30.

P(neither) = 7/30 ≈ 0.233
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Step 5 — Verify

Check: 11/30 + 7/30 + 5/30 + 7/30 = 30/30 = 1. ✓ All regions sum to the total, confirming the answer.

SECTION 7

Venn Diagrams vs. Tree Diagrams — When to Use Which

Both Venn diagrams and tree diagrams are representations of probability, but they excel in different situations. Knowing which tool to reach for can save you valuable time on the IB exam.

Feature	Venn Diagram	Tree Diagram
Best for	Overlapping events from a single experiment	Sequential (multi-stage) experiments
Shows intersections	Yes — overlap region is clearly visible	Indirectly — combine branch paths
Shows conditional probabilities	Not directly	Yes — second-stage branches are conditional
Number of events	Practical for 2 or 3 events	Handles any number of stages
Replacement matters?	Not typically shown	Easily captured by changing branch probabilities

Comparison of Venn diagrams and tree diagrams for probability problems

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KEY TAKEAWAY

Think of a Venn diagram as a snapshot — it captures how groups overlap at a single moment. A tree diagram is more like a flowchart — it walks you through decisions or events in order, step by step. If the problem says "and then," reach for a tree. If it says "how many belong to both," reach for a Venn.

SECTION 8

Connection to Conditional Probability & Beyond

The ideas in SL 4.3 are the foundation for everything that comes next in IB probability. Once you are comfortable with sample spaces, events, and the addition rule, you are ready for conditional probability (SL 4.4) and independent events (SL 4.5). In fact, tree diagrams already hint at conditional probability: the second-stage branch probabilities depend on what happened at the first stage.

Concept	SL 4.3 (This Lesson)	SL 4.4–4.5 (Next Steps)
Sample space	List or diagram all outcomes	Restricted sample spaces (given conditions)
Intersection	A ∩ B from Venn diagrams	P(A ∩ B) = P(A) × P(B \| A)
Addition rule	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	Extended to conditional scenarios
Tree diagrams	Multiply along paths, add across paths	Bayes' theorem applications

How SL 4.3 concepts evolve into more advanced probability topics

If you are planning to take the HL course, these fundamentals also feed into probability distributions (HL 4.12–4.14), where you model the number of times an event occurs over many trials. For now, focus on building a rock-solid understanding of sample spaces, the addition rule, and diagram skills — everything else builds on these.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A standard six-sided die is rolled once. Write down the sample space U. Then let A = {rolling a prime number}. List the elements of A and find P(A). Is event A the same as event A′? Explain.

Show answer

The sample space is U = {1, 2, 3, 4, 5, 6}. The prime numbers in this set are 2, 3, and 5, so A = {2, 3, 5}. Since there are 3 favourable outcomes out of 6 equally likely outcomes, P(A) = 3/6 = 1/2. The complement A′ = {1, 4, 6} contains the non-prime outcomes, and P(A′) = 1 − 1/2 = 1/2. Even though P(A) = P(A′) here, A and A′ are not the same event — they contain entirely different outcomes.

Question 2: BASIC CALCULATION

In a group of 40 students, 25 play football and 15 play basketball. 8 play both. Find P(a randomly selected student plays football or basketball).

Show answer

Using the addition rule: P(F ∪ B) = P(F) + P(B) − P(F ∩ B) = 25/40 + 15/40 − 8/40 = 32/40 = 4/5 = 0.8. You can verify by counting: only football = 25 − 8 = 17, only basketball = 15 − 8 = 7, both = 8, so total who play at least one sport = 17 + 7 + 8 = 32. Then 32/40 = 0.8. ✓

Question 3: INTERMEDIATE

A bag contains 4 green and 6 yellow balls. Two balls are drawn one after the other without replacement. Draw a tree diagram and find the probability that both balls are the same colour.

Show answer

Stage 1: P(G₁) = 4/10, P(Y₁) = 6/10. Stage 2 after Green: P(G₂) = 3/9, P(Y₂) = 6/9. Stage 2 after Yellow: P(G₂) = 4/9, P(Y₂) = 5/9. Path GG = (4/10)(3/9) = 12/90. Path YY = (6/10)(5/9) = 30/90. P(same colour) = 12/90 + 30/90 = 42/90 = 7/15 ≈ 0.467. You can check: all four paths sum to 12/90 + 24/90 + 24/90 + 30/90 = 90/90 = 1. ✓

Question 4: APPLIED

A school survey finds that P(student owns a laptop) = 0.72, P(student owns a tablet) = 0.45, and P(student owns both) = 0.30. A student is chosen at random. Find the probability that the student owns neither a laptop nor a tablet, and explain what this means in context.

Show answer

First, apply the addition rule: P(L ∪ T) = 0.72 + 0.45 − 0.30 = 0.87. Then use the complement: P(neither) = 1 − 0.87 = 0.13. This means that 13% of students in the school do not own a laptop or a tablet. On a Venn diagram, these 13% sit outside both circles, inside the rectangle. Knowing this figure could help the school assess how many students might need access to shared devices.

Question 5: CRITICAL THINKING

Events A and B are such that P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7. Determine P(A ∩ B). Are A and B mutually exclusive? Could A and B be complementary events? Justify your answers.

Show answer

Rearranging the addition rule: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2. Since P(A ∩ B) = 0.2 ≠ 0, the events are not mutually exclusive — there is a non-zero chance both occur simultaneously. For A and B to be complementary, we would need P(A) + P(B) = 1 (which gives 0.5 + 0.4 = 0.9 ≠ 1) and A ∩ B = ∅ (which we just showed is false). Therefore A and B are not complementary. This problem illustrates that overlapping events can still have a union less than 1, leaving room for outcomes in neither event; here P(neither) = 1 − 0.7 = 0.3.

SUMMARY

Lesson Summary

Summary

Probability measures how likely an event is to occur, always between 0 and 1. The sample space U is the set of all possible outcomes, and an event is any subset of U. For equally likely outcomes, P(A) = n(A) / n(U). The complement rule P(A′) = 1 − P(A) lets you find the probability that an event does not occur, which is often the faster approach.

Venn diagrams visualize overlapping events with circles, making the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B) intuitive — subtract the overlap to avoid double-counting. Tree diagrams handle sequential events by multiplying along branches and adding across paths, naturally accommodating with-replacement and without-replacement scenarios. Together, these tools give you a complete toolkit for tackling any probability question at the SL level.