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Probability Unit 2: Axioms of Probability
Probability is built on a sample space of outcomes, events that are subsets of it, and three axioms. Every rule you use later, from the complement rule to inclusion–exclusion, follows from these axioms.
Covers: Sample spaces and events, Set operations and De Morgan's laws, Kolmogorov's axioms, Complement and addition rules, Equally likely outcomes, Inclusion–exclusion for probabilities, Geometric probability
Study guide
An experiment has a sample space S, the set of all possible outcomes, and an event is any subset of S. Events combine like sets: A ∪ B means A or B (or both) occurs, A ∩ B means both occur, and Aᶜ means A does not occur. Two events are mutually exclusive, or disjoint, when A ∩ B = ∅. De Morgan's laws, (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ and (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ, turn 'neither' and 'not both' statements into forms you can compute. Drawing a Venn diagram and labeling each region is often the quickest way to solve a word problem about overlapping groups.
Kolmogorov's axioms say three things: P(A) ≥ 0 for every event, P(S) = 1, and if A₁, A₂, … are mutually exclusive, the probability of their union is the sum of their probabilities. From these come the rules you use daily. The complement rule P(Aᶜ) = 1 − P(A) holds because A and Aᶜ split S. If A ⊆ B, then P(A) ≤ P(B). The addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B) subtracts the overlap that would otherwise be counted twice, and its three-event version adds back the triple intersection. These rules also show which assignments are impossible, such as an intersection more likely than one of its events.
When a sample space is finite and all outcomes are equally likely, P(A) = (number of outcomes in A)/(number in S), so the counting tools from Unit 1 become probability tools. The trap is assuming equal likelihood where it does not hold: the 11 possible sums of two dice are not equally likely, but the 36 ordered pairs are. For 'at least one' events, compute the complement 'none'. The birthday problem is the classic example: the chance that 23 people all have different birthdays is below ½, so a shared birthday is more likely than not.
Some problems use continuous sample spaces. If a point is chosen uniformly from a region, the probability of an event is its area divided by the total area; this is geometric probability. Probability also has a continuity property: for an increasing sequence of events, the probability of their union is the limit of their probabilities, and similarly for decreasing sequences and intersections. On exams, many questions give a few probabilities about two or three events and ask for another. Write down what is given, decide which region of the Venn diagram is wanted, and solve with the addition and complement rules.
Key ideas
- Events are subsets of the sample space and combine with union, intersection and complement.
- The three axioms are nonnegativity, P(S) = 1 and additivity for disjoint events.
- The complement and addition rules follow directly from the axioms.
- Equally likely outcomes reduce probability to counting favorable over total.
- Disjoint is not the same as independent; disjoint events with positive probability are never independent.
- For 'at least one' events, 1 − P(none) is usually simplest.
- For uniform points in a region, probability is a ratio of areas.
Formulas and rules
Common mistakes
- Adding P(A) + P(B) for events that overlap, which double-counts A ∩ B.
- Multiplying probabilities as if events were independent when nothing says they are.
- Treating outcomes such as dice sums as equally likely when they are not.
- Confusing 'exactly one' with 'at least one' (the union).
- Mixing up the two De Morgan laws when finding 'neither' or 'not both'.
Before the test, make sure you can
- State Kolmogorov's three axioms and derive the complement rule.
- Translate words such as 'neither', 'exactly one' and 'not both' into set notation.
- Calculate P(A ∪ B) and P(A ∩ B) from the addition rule.
- Apply three-event inclusion–exclusion.
- Compute equally likely probabilities with counting, including cards and dice.
- Use the complement for 'at least one' problems such as the birthday problem.
- Identify impossible probability assignments.
- Compute geometric probabilities as ratios of areas.
Words to know
- Sample space
- The set S of all possible outcomes of an experiment.
- Event
- A subset of the sample space.
- Union A ∪ B
- The event that A occurs, B occurs, or both.
- Intersection A ∩ B
- The event that both A and B occur.
- Complement Aᶜ
- The event that A does not occur.
- Mutually exclusive
- Events with no outcomes in common, so A ∩ B = ∅.
- Kolmogorov's axioms
- P(A) ≥ 0, P(S) = 1, and the probability of a union of disjoint events is the sum of their probabilities.
- Complement rule
- P(Aᶜ) = 1 − P(A).
- Addition rule
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- De Morgan's laws
- (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ and (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ.
- Equally likely outcomes
- A model in which every outcome in a finite S has probability 1/|S|.
- Monotonicity
- If A ⊆ B, then P(A) ≤ P(B).
- Partition
- A collection of disjoint events whose union is S.
- Geometric probability
- For a uniform point in a region, probability equals the event's area over the total area.
- Continuity of probability
- For increasing events, P(∪Aₙ) = lim P(Aₙ).
Sample practice questions
30 more questions like these, from level 1 to 5, are in the practice set.
Which statement is one of Kolmogorov's axioms of probability?
- Every event other than S has probability at most ½
- For any two events, P(A ∪ B) = P(A) + P(B)
- For any two events, P(A ∩ B) = P(A)P(B)
- If A₁, A₂, … are mutually exclusive, P(A₁ ∪ A₂ ∪ ···) = P(A₁) + P(A₂) + ···
Show the answer and steps
Answer: D) If A₁, A₂, … are mutually exclusive, P(A₁ ∪ A₂ ∪ ···) = P(A₁) + P(A₂) + ···
- The axioms are: P(A) ≥ 0, P(S) = 1, and countable additivity for disjoint events.
- Additivity holds only for mutually exclusive events.
- P(A ∩ B) = P(A)P(B) defines independence, which is not an axiom.
What does the complement rule state?
- P(Aᶜ) = 1 − P(A)
- P(Aᶜ) = P(S) + P(A)
- P(Aᶜ) = 1/P(A)
- P(Aᶜ) = P(A) − 1
Show the answer and steps
Answer: A) P(Aᶜ) = 1 − P(A)
- A and Aᶜ are disjoint and together make up S.
- So P(A) + P(Aᶜ) = P(S) = 1.
- Solving gives P(Aᶜ) = 1 − P(A).
Two fair dice are rolled. What is the probability that at least one die shows a 6?
- 11/36
- 1/6
- 5/18
- 1/3
Show the answer and steps
Answer: A) 11/36
- P(no 6) = (5/6)² = 25/36.
- P(at least one 6) = 1 − 25/36 = 11/36.
- 1/3 = 1/6 + 1/6 double-counts the outcome (6, 6).