Probability is the mathematical quantification of uncertainty. It assigns a numerical value to the likelihood that a particular outcome or event will occur as the result of a random experiment.

The set of all possible outcomes of an experiment is called the sample space, denoted by S (or the universal set ξ\xi). An event is any subset of the sample space. The probability of an event A is denoted by P(A).

Illustration of the universal sample space with overlapping probability circles
Image Source: Gianpiero Placidi

Why Probability Properties Matter

  • Real-World Applications: Probability theory governs actuarial science (life expectancy and insurance premium calculations), financial risk modelling, predictive weather forecasting, clinical drug trial analysis, and machine learning classification algorithms.
  • Academic Relevance: The axiomatic properties of probability provide the rigorous theoretical framework for discrete and continuous probability distributions (Binomial, Poisson, Normal), random variables, expectations, and hypothesis testing.
The best Maths tutors available
Poonam
5
5 (80 reviews)
Poonam
£100
/h
Gift icon
1st lesson free!
Paolo
4.9
4.9 (85 reviews)
Paolo
£35
/h
Gift icon
1st lesson free!
Anthony
5
5 (95 reviews)
Anthony
£10
/h
Gift icon
1st lesson free!
Sehaj
4.9
4.9 (62 reviews)
Sehaj
£60
/h
Gift icon
1st lesson free!
Intasar
5
5 (73 reviews)
Intasar
£99
/h
Gift icon
1st lesson free!
Shane
4.9
4.9 (51 reviews)
Shane
£50
/h
Gift icon
1st lesson free!
Dr scott
5
5 (78 reviews)
Dr scott
£100
/h
Gift icon
1st lesson free!
Tomi
5
5 (41 reviews)
Tomi
£55
/h
Gift icon
1st lesson free!
Poonam
5
5 (80 reviews)
Poonam
£100
/h
Gift icon
1st lesson free!
Paolo
4.9
4.9 (85 reviews)
Paolo
£35
/h
Gift icon
1st lesson free!
Anthony
5
5 (95 reviews)
Anthony
£10
/h
Gift icon
1st lesson free!
Sehaj
4.9
4.9 (62 reviews)
Sehaj
£60
/h
Gift icon
1st lesson free!
Intasar
5
5 (73 reviews)
Intasar
£99
/h
Gift icon
1st lesson free!
Shane
4.9
4.9 (51 reviews)
Shane
£50
/h
Gift icon
1st lesson free!
Dr scott
5
5 (78 reviews)
Dr scott
£100
/h
Gift icon
1st lesson free!
Tomi
5
5 (41 reviews)
Tomi
£55
/h
Gift icon
1st lesson free!
Let's go

Core Theory

1. Kolmogorov's Axioms of Probability

Modern probability theory is built upon three foundational axioms established by mathematician Andrey Kolmogorov:

Axiom 1: Non-Negativity

The probability of any event A is a non-negative real number:

P(A)≥0for every event A⊆SP(A) \ge 0 \quad \text{for every event } A \subseteq S

Axiom 2: Total Probability of the Sample Space

The probability that at least one of the elementary outcomes in the entire sample space S will occur is exactly 1 (certainty):

P(S)=1P(S) = 1

Combining Axioms 1 and 2 defines the fundamental probability scale:

0≤P(A)≤10 \le P(A) \le 1
  • P(A) = 0: An impossible event (it can never happen).
  • P(A) = 1: A certain event (it is guaranteed to happen).

Axiom 3: Additivity for Mutually Exclusive Events

If two events A and B cannot occur at the same time (they are disjoint or mutually exclusive, meaning A∩B=∅A \cap B = \emptyset), the probability of either event occurring is the sum of their individual probabilities:

P(A∪B)=P(A)+P(B)if A∩B=∅P(A \cup B) = P(A) + P(B) \quad \text{if } A \cap B = \emptyset

2. Fundamental Derived Properties

From Kolmogorov's three axioms, all other probability rules are logically derived:

A. The Complementary Event Rule

The complement of event A, denoted by A′A' (or AcA^c or A‾\bar{A}), is the event that A does not occur.

Because an event and its complement partition the sample space into two disjoint subsets (A∪A′=SA \cup A' = S and A∩A′=∅A \cap A' = \emptyset):

P(A)+P(A′)=P(S)=1P(A) + P(A') = P(S) = 1

P(A′)=1−P(A)⟺P(A)=1−P(A′)P(A') = 1 - P(A) \iff P(A) = 1 - P(A')

B. The Impossible Event

The empty set ∅\emptyset represents an event containing zero outcomes:

P(∅)=0P(\emptyset) = 0

C. Subset Monotonicity

If event A is a subset of event B (A⊆BA \subseteq B), then whenever A occurs, B must also occur. Consequently:

P(A)≤P(B)P(A) \le P(B)

P(B∖A)=P(B∩A′)=P(B)−P(A)P(B \setminus A) = P(B \cap A') = P(B) - P(A)

3. The Addition Law (Union of Events)

The union of two events, denoted by A∪BA \cup B, represents the event that A occurs, B occurs, or both occur (inclusive "or").

Illustration of the addition law showing mutually and non-mutually exclusive events
Image Source: Gianpiero Placidi

The General Addition Law:

When A and B can occur simultaneously, adding P(A) and P(B) directly counts the shared intersection (A∩BA \cap B) twice. We must subtract this overlap once:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Mutually Exclusive Case:

If events A and B are mutually exclusive, they have no shared outcomes (P(A∩B)=0P(A \cap B) = 0):

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

Extension to Three Events:

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

4. Conditional Probability & The Multiplication Law

The conditional probability of event A given that event B has already occurred is denoted by P(A|B)P(A\vert{}B) (read as "the probability of A given B").

Conditioning on B restricts the operational sample space from the entire universe S down to the subset B:

P(A|B)=P(A∩B)P(B)(provided P(B)>0)P(A\vert{}B) = \dfrac{P(A \cap B)}{P(B)} \quad (\text{provided } P(B) > 0)

Similarly:

P(B|A)=P(A∩B)P(A)(provided P(A)>0)P(B\vert{}A) = \dfrac{P(A \cap B)}{P(A)} \quad (\text{provided } P(A) > 0)

The General Multiplication Law:

Rearranging the conditional probability formula gives the probability of the intersection (both events occurring together):

P(A∩B)=P(B)×P(A|B)=P(A)×P(B|A)P(A \cap B) = P(B) \times P(A\vert{}B) = P(A) \times P(B\vert{}A)

5. Independent Events vs. Mutually Exclusive Events

  • Independent Events: The occurrence of one event has no influence on the probability of the other:

P(A∩B)=P(A)×P(B)P(A ∩ B) = P(A) × P(B)

P(A|B)=P(A)P(A|B) = P(A)
  • Mutually Exclusive Events: The two events cannot happen at the same time (no overlap)

P(A∩B)=0P(A ∩ B) = 0

P(A∪B)=P(A)+P(B)P(A ∪ B) = P(A) + P(B)

Worked Examples

1

Worked Example 1: Basic Axioms and Complementary Events

A biased four-sided spinner has faces labeled 1, 2, 3, and 4. The probabilities of landing on faces 1, 2, and 3 are given as:

(a) Calculate the probability of landing on face 4.

(b) Find the probability of landing on an even number.

(c) Find the probability of not landing on face 2.

Solution

Step 1: Calculate P(4) using the total probability axiom

The sum of all mutually exclusive elementary outcomes in the sample space must equal 1:

Substitute the known values:

Step 2: Calculate the probability of landing on an even number

The even outcomes are 2 and 4. Because these outcomes cannot occur simultaneously, they are mutually exclusive:

Step 3: Calculate the probability of not landing on face 2

Apply the complementary event rule:

2

Worked Example 2: The General Addition Law and Venn Diagram Partitions

In a sixth-form cohort of 120 students, events A and B represent participation in drama club and choir, respectively. The probabilities are:

(a) Find the probability that a randomly chosen student participates in both activities, .

(b) Calculate the probability that a student participates in drama club only, .

(c) Calculate the probability that a student participates in neither activity, .

Solution

Step 1: Rearrange the general addition law to solve for the intersection

The general addition law for any two non-mutually exclusive events is:

Rearranging gives:

Step 2: Calculate the intersection probability

Substitute the given values:

Step 3: Calculate the probability that only event A occurs

The region for event A consists of the exclusive part of A and the shared overlap:

Step 4: Calculate the probability that neither event occurs

By de Morgan's laws, the event that neither A nor B occurs is the complement of their union:

Apply the complement rule:

3

Worked Example 3: Conditional Probability and the Multiplication Rule

In a clinical screening trial, 5% of a population carries a specific medical condition (event D). A diagnostic test produces a positive result (event T) according to the following specifications:

If a patient has the condition, the test is positive with probability 0.92:

If a patient does not have the condition, the test produces a false positive with probability 0.04:

(a) Calculate the probability that a randomly chosen individual has the condition and tests positive, .

(b) Calculate the overall probability that a randomly selected person tests positive, .

(c) Find the conditional probability that a person who tests positive actually has the condition, .

Solution

Step 1: Calculate the joint probability for a true positive

Using the multiplication rule:

Step 2: Calculate the joint probability for a false positive

First, find the probability that a patient does not have the condition:

Now apply the multiplication rule for false positives:

Step 3: Calculate the total probability of testing positive

Because having the condition and not having the condition are mutually exclusive and exhaustive:

Step 4: Calculate the revised conditional probability

Apply the definition of conditional probability:

The probability that a person testing positive actually has the condition is approximately 54.76%.

4

Worked Example 4: Testing for Independence and Mutual Exclusivity
Two events A and B have probabilities:

(a) Determine whether A and B are mutually exclusive.

(b) Determine whether A and B are statistically independent.

(c) Evaluate .

Solution

Step 1: Calculate the intersection probability

Use the general addition rule:

Step 2: Test for mutual exclusivity

Two events are mutually exclusive if and only if their intersection is impossible:

Since , the two events can occur simultaneously.

Therefore, events A and B are not mutually exclusive.

Step 3: Test for statistical independence

Two events are statistically independent if and only if the product of their individual probabilities equals their intersection:

Because , the independence condition is satisfied.

Therefore, events A and B are statistically independent.

Step 4: Evaluate the conditional probability

Because events A and B are independent, the occurrence or non-occurrence of event B has no effect on the likelihood of event A:

Algebraic verification:

5

Worked Example 5: Solving Algebraic Equations Using Probability Axioms
Two events A and B have probabilities expressed in terms of an unknown constant p:

(a) Given that A and B are mutually exclusive and , find the value of p.

(b) Given instead that A and B are independent and :

Show that .
Find the value of p, explaining why one mathematical solution must be rejected.

Solution

Worked Example 5: Solving Algebraic Equations Using Probability Axioms
Two events A and B have probabilities expressed in terms of an unknown constant p:

(a) Given that A and B are mutually exclusive and , find the value of p.

(b) Given instead that A and B are independent and :

Show that .
Find the value of p, explaining why one mathematical solution must be rejected.

Summarise with AI:

Did you like this article? Rate it!

4.00 (5 rating(s))
Loading...

Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.