Probability Axioms for Finite Models

From Outcomes to Events

A finite experiment has a list of possible outcomes, but a probability model must also describe how likely those outcomes are. Assigning probabilities to individual outcomes gives a systematic way to find the probability of any event, where an event is a set of outcomes. The assignment must satisfy precise rules: a collection of numbers that fails those rules is not a probability model.

For example, a hypothetical device might produce one of several labeled results, and a quality-control question might ask for the probability that the result belongs to a specified group. The calculation depends on the outcome probabilities, not just on how many results are in the group. Equal probabilities are appropriate only when the model says the outcomes are equally likely.

Definition: A finite sample space $ \Omega $ is a finite set of mutually exclusive and collectively exhaustive outcomes for an experiment. An event $A$ is any subset of $ \Omega $. A probability model assigns a number $P(A)$ to each event, subject to the probability axioms.

The Probability Axioms

For a finite sample space, every subset of $ \Omega $ is an event. A probability assignment must satisfy the following axioms. In the third axiom, the events in the sequence must be pairwise disjoint, meaning that no outcome belongs to more than one of them.

Probability axioms:

Nonnegativity: For every event $A$, $P(A) \geq 0$.
Normalization: $P(\Omega)=1$.
Countable additivity: For any pairwise disjoint events $A_1,A_2,\ldots$, $P\left(\bigcup_{j=1}^{\infty} A_j\right)=\sum_{j=1}^{\infty}P(A_j)$.

For a finite sample space, the third axiom reduces to adding the probabilities of finitely many disjoint events. The empty event has probability zero.

The individual outcomes are often called the atoms of a finite model. Write $p(\omega)=P(\{\omega\})$ for the probability assigned to the single-outcome event containing $ \omega $. The entire model can then be specified by giving one number $p(\omega)$ for each outcome.

Finite-model rule: The outcome probabilities must satisfy $p(\omega)\geq 0$ for every $\omega\in\Omega$, and $$ \sum_{\omega\in\Omega}p(\omega)=1. $$ For an event $A\subseteq\Omega$, $$ P(A)=\sum_{\omega\in A}p(\omega). $$

The axioms and this rule are two ways to describe the same finite model. The outcome probabilities are nonnegative and add to one; the probability of an event is the sum of the probabilities of the outcomes it contains. No assumption that outcomes are equally likely is built into the axioms.

Why Event Probabilities Are Sums

Every event $A$ in a finite sample space can be written as a union of the single-outcome events $ \{\omega\} $ for outcomes $ \omega\in A $. These single-outcome events are disjoint: an outcome cannot be two different elements of the sample space at once. Applying additivity to that disjoint union gives the event-sum formula.

A=\bigcup_{\omega\in A}\{\omega\}, \qquad P(A)=\sum_{\omega\in A}P(\{\omega\}) =\sum_{\omega\in A}p(\omega).

This argument also explains why a list of outcome probabilities determines the probability of every event. Once the probabilities of the single-outcome events are specified, additivity fixes the probability of any union of them. For the whole sample space, the sum of all outcome probabilities must be one because $P(\Omega)=1$.

Checking and Using a Finite Probability Model

A reliable calculation separates model validation from event calculation. First verify that the outcomes are the complete list of possibilities and that the assigned numbers are nonnegative and sum to one. Then identify exactly which outcomes constitute the requested event and add only their probabilities.

List the outcomes.
Use a finite sample space whose outcomes are mutually exclusive and collectively exhaustive for the experiment described.

Check the assignments.
Confirm that each outcome probability is nonnegative and that the sum across all outcomes is one.

Specify the event.
Translate the event in words into a set of outcomes, taking care to include every outcome that meets the condition.

Add the event's outcome probabilities.
Use $P(A)=\sum_{\omega\in A}p(\omega)$, not a count of outcomes unless equal likelihood has been established.

Worked Example: A Valid Model With Unequal Outcome Probabilities

A hypothetical sorting device reports one of four labels: $R$, $S$, $T$, or $U$. Its calibrated finite model assigns probabilities $0.10$, $0.25$, $0.40$, and $0.25$, respectively. Find the probability that the device reports either $S$ or $U$. The model assumption is that these four labels are all possible outcomes and that the stated probabilities apply to one use of the device.

Method: validate the outcome probabilities, then sum over the event. First check nonnegativity: each of $0.10$, $0.25$, $0.40$, and $0.25$ is nonnegative. Next check normalization:

$$ 0.10+0.25+0.40+0.25=1.00. $$

Thus the supplied values satisfy the finite-model conditions. The event is $A=\{S,U\}$, so its probability is the sum of the probabilities assigned to those two outcomes:

$$ P(A)=p(S)+p(U)=0.25+0.25=0.50. $$

The answer is $0.50$. A common mistake would be to count two favorable labels out of four and report $2/4=0.50$ on the grounds that the labels are equally likely. That reasoning is not justified by the model: the outcomes have unequal probabilities. The numerical answer happens to match in this example, but the correct method is to add the stated probabilities.

Worked Example: Converting Relative Weights to Probabilities

A hypothetical software test produces one of three mutually exclusive outcomes: “pass,” “review,” or “fail.” The test model gives relative weights $2$, $3$, and $5$, respectively. Treat these as relative weights, not as probabilities, and calculate the probability of “review” or “fail.” Assume these are the only possible outcomes.

Method: normalize the weights, then add the probabilities in the event. The total weight is

$$ 2+3+5=10. $$

Divide each weight by the total to obtain the outcome probabilities:

p(\text{pass})=\frac{2}{10}=0.20,\qquad p(\text{review})=\frac{3}{10}=0.30,\qquad p(\text{fail})=\frac{5}{10}=0.50.

They are nonnegative and sum to $0.20+0.30+0.50=1.00$, as required. The requested event is $\{\text{review},\text{fail}\}$, so

P(\text{review or fail})=0.30+0.50=0.80.

The answer is $0.80$. The common mistake is to treat the weight $3$ as the probability of “review,” or to add the raw weights and call $3+5=8$ a probability. Relative weights must first be divided by their total.

Worked Example: An Event in a Two-Attribute Outcome Space

A hypothetical inspection program records both a zone, $N$ or $S$, and a status, $C$ or $D$. Assume exactly one zone and one status are recorded, so the four ordered pairs form the complete finite sample space. The model assigns the following probabilities:

p(N,C)=0.12,\quad p(N,D)=0.18,\quad p(S,C)=0.08,\quad p(S,D)=0.62.

Find the probability that the inspection is in zone $N$ with status $D$, or in zone $S$ with status $C$. These are the only two outcomes in the requested event.

Method: represent the event as a subset of the ordered-pair sample space. Before calculating, check that the four outcome probabilities are nonnegative and sum to one:

$$ 0.12+0.18+0.08+0.62=1.00. $$

The event is $A=\{(N,D),(S,C)\}$. Its probability is therefore

$$ P(A)=p(N,D)+p(S,C)=0.18+0.08=0.26. $$

The answer is $0.26$. The common mistake is to include $(N,C)$ merely because it has zone $N$, or $(S,D)$ merely because it has zone $S$. The event requires the specified zone-and-status pairs; list those pairs before adding probabilities.

Common Mistakes and Why the Model Matters

A probability calculation has two parts: the model and the arithmetic. Correct addition cannot repair an invalid model, and a valid model does not make every event equally likely. In particular, the number of outcomes in an event is not enough to determine its probability unless the outcomes are known to have equal probabilities.

Using negative outcome probabilities. An assignment with any negative value violates nonnegativity, even if all the values sum to one.
Using probabilities that do not sum to one. The total must equal one across the complete sample space. If the values are described as relative weights instead, normalize them before treating them as probabilities.
Assuming equal likelihood from the way outcomes are listed. A list of four outcomes does not imply probability $1/4$ for each. Equal likelihood must be part of the model.
Adding the wrong outcomes. Translate the event into a set before calculating. For outcomes with several attributes, preserve all the conditions in the event description.

Key takeaway: In a finite probability model, assign each outcome a nonnegative probability, require the probabilities over the complete sample space to sum to one, and find an event's probability by summing the probabilities of the outcomes in that event.

Check Your Understanding

For each problem, first identify the outcome probabilities that are relevant to the requested event. Where needed, check that the proposed model is valid.

A hypothetical scanner has outcomes $L$, $M$, and $H$, with probabilities $0.20$, $0.50$, and $0.30$. What is the probability of an outcome in $\{L,H\}$? (Answer: $0.50$)
A finite model assigns probabilities $0.15$, $0.35$, and $0.45$ to its three listed outcomes. Is this a valid probability model? (Answer: No; the probabilities sum to $0.95$, not $1$.)
A hypothetical claims-routing rule gives outcomes $A$, $B$, and $C$ relative weights $1$, $4$, and $5$. After normalizing, what is the probability of outcome $B$ or $C$? (Answer: $0.90$)
A hypothetical device produces ordered pairs $(1,X),(1,Y),(2,X),(2,Y)$, with probabilities $0.10$, $0.20$, $0.30$, and $0.40$, in that order. Find the probability of $(1,Y)$ or $(2,X)$. (Answer: $0.50$)
A model lists four mutually exclusive and exhaustive outcomes with probabilities $0.12$, $0.28$, $0.22$, and $0.38$. An event consists of the first and fourth outcomes. Calculate its probability and state whether the model is valid. (Answer: The model is valid because the probabilities sum to $1.00$; the event probability is $0.50$.)

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