Experiments, Outcomes, and Probability Models

From a Physical Procedure to a Probability Model

A coin toss takes place in the physical world: a particular coin is released, moves through the air, and lands. A probability calculation does not track all of that motion. Instead, it uses a simplified description of what will be recorded, such as “heads” or “tails,” together with probabilities for those possible results. Separating the procedure, the recorded result, and the mathematical model prevents a common confusion: a model describes uncertainty about an experiment; it is not the experiment itself.

This distinction matters in statistics because probability statements always depend on a description of what can be observed and on assumptions about how the procedure operates. Changing the recorded result or the assumptions can change the model, even when the physical setting appears similar.

Definition: A random experiment is a specified procedure that produces a recorded result whose value is not known with certainty in advance. An outcome is one possible value of that recorded result. The word “random” describes uncertainty in the procedure’s result, not an absence of physical causes.

The recording convention is part of the description. If a card is drawn, the recorded outcome might be its rank, its suit, or the exact card. These are different possible descriptions of the same physical draw. The outcome is the value specified by the recording convention, not every physical detail of what occurred.

A mathematical probability model represents the possible recorded outcomes and assigns probabilities to them. For the finite examples in this tutorial, it is enough to list the outcomes and give each a nonnegative probability, with all the probabilities adding to one. More general probability models use additional mathematical structure, developed later.

Definition: A finite probability model consists of a finite list \(\Omega\) of possible outcomes and a probability assignment \(p(\omega)\) for each \(\omega\in\Omega\), satisfying \(p(\omega)\geq 0\) and \(\sum_{\omega\in\Omega}p(\omega)=1\). The list \(\Omega\) is the model’s outcome space. A probability assigned by the model describes the chance of the corresponding outcome under the assumptions used to build the model.

A model is useful only relative to a stated procedure and recording rule. For example, a model of a coin toss that records heads and tails does not describe the coin’s landing position, spin rate, or distance traveled. Those details are omitted because they are not part of the recorded result being modeled.

Basic Consequences of a Finite Model

The total probability assigned to the outcome list is one by definition. Two simple consequences help interpret this condition and check calculations.

Theorem: If a finite model has \(n\) outcomes and assigns the same probability \(c\) to each, then \(c=1/n\).

Proof. Since the model’s outcome probabilities sum to one and each of the \(n\) outcomes has probability \(c\), their sum is \(nc\). Thus \(nc=1\), so \(c=1/n\). This conclusion applies when equal probabilities are part of the model; it does not establish that the outcomes in a real procedure are equally likely. \(\square\)

A procedure may also record a coarser result than the underlying outcome description. For instance, it may record only whether a drawn card is a face card, rather than its exact identity. The probabilities of the coarser recorded values are obtained by grouping the probabilities of the detailed outcomes that produce each value.

Result: Suppose \(g\) maps each outcome \(\omega\) in a finite model to a recorded value \(g(\omega)\). The probability assigned to a recorded value \(y\) is \(q(y)=\sum_{\omega:g(\omega)=y}p(\omega)\). The resulting probabilities \(q(y)\) are nonnegative and sum to one over the recorded values that can occur.

Derivation. Each \(q(y)\) is a sum of nonnegative terms, so it is nonnegative. Every original outcome \(\omega\) maps to exactly one recorded value \(g(\omega)\). Therefore, when the sums defining \(q(y)\) are added over all possible recorded values, each original probability \(p(\omega)\) appears exactly once. Hence \(\sum_y q(y)=\sum_{\omega\in\Omega}p(\omega)=1\). So the coarser recording also has a valid probability assignment. \(\square\)

Worked Models

Worked Example: A Coin Toss

A coin is tossed once, and the recorded result is whether it lands heads or tails. Suppose the model treats the coin as fair. Specify the model and calculate the probability of each possible result.

The recording rule gives two outcomes: \(H\) for heads and \(T\) for tails. Thus \(\Omega=\{H,T\}\). The fair-coin assumption assigns probability \(1/2\) to each outcome: \(p(H)=1/2\) and \(p(T)=1/2\).

The normalization check is \(p(H)+p(T)=1/2+1/2=1\), so these assignments form a probability model. The model says that heads and tails each have probability \(1/2\) under the fair-coin assumption. It does not state that every physical toss must give one of the two outcomes in a perfectly regular pattern, nor does it predict a particular toss.

If the physical coin is biased, the same recording rule still gives \(H\) and \(T\), but the probability assignment may differ. For example, if the model assigns \(p(H)=0.7\), then normalization requires \(p(T)=1-0.7=0.3\). The possible outcomes have not changed; the model’s assumptions about their probabilities have.

Worked Example: Recording a Card’s Rank

One card is drawn from a thoroughly shuffled standard 52-card deck, and only its rank is recorded. Use the model that assigns equal probability to each physical card to find the probability that the recorded rank is an ace.

The possible recorded ranks are ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king. The model’s detailed outcomes are the 52 individual cards. By the stated assumption, each card has probability \(1/52\).

There are four aces, one in each suit. Each of those four detailed outcomes produces the recorded value “ace.” The coarser-recording rule therefore gives \(q(\text{ace})=4(1/52)=4/52=1/13\).

As a check, each of the 13 ranks is represented by four cards, so each rank receives probability \(4/52=1/13\). The total over all recorded ranks is \(13(1/13)=1\). This calculation depends on the model’s assumption about the shuffle and the detailed card outcomes; “a card is drawn” alone does not supply those probability assignments.

Worked Example: Inspecting a Component

A component is selected at random from a lot containing 200 components, of which 10 are defective. An inspection records only “defective” or “not defective,” and assume the inspection classifies the selected component correctly. Find the probability model for the recorded result.

The physical procedure is selecting one component and inspecting it. The recorded outcomes are \(D\), meaning defective, and \(N\), meaning not defective. Under the assumption that each of the 200 components is equally likely to be selected, there are 10 component outcomes producing \(D\), so \(p(D)=10/200=0.05\).

The other \(200-10=190\) components produce \(N\), giving \(p(N)=190/200=0.95\). The probabilities are nonnegative and sum to one: \(0.05+0.95=1\). Thus the model assigns probability \(0.05\) to a defective inspection result and \(0.95\) to a not-defective result.

The inspection assumption is important. If the inspection could misclassify components, the recorded result would not necessarily match the component’s true condition, and this probability model would need to represent that additional uncertainty. The physical lot, the inspection procedure, and the recorded label are distinct ingredients.

What a Model Does—and Does Not—Say

A probability model is a mathematical description built for a purpose. It makes assumptions explicit, lets us calculate probabilities from them, and provides a common language for reasoning about uncertain results. Its conclusions are conditional on the model: if the coin is not fair, the fair-coin probabilities may not describe it; if cards are not shuffled adequately, equal probabilities for the cards may not be appropriate.

The model is not a claim that an outcome is guaranteed to occur in proportion to its assigned probability in every short run. In the coin model, \(p(H)=1/2\) describes the assigned chance of heads on a toss under the model’s assumptions. It does not mean that every pair of tosses contains exactly one head. Likewise, the inspection model’s \(0.05\) is based on selecting from the stated lot under the stated selection assumption; it is not a statement about every lot or every inspection process.

Common Mistakes

Confusing the procedure with the outcome. Tossing a coin is the experiment; “heads” is one possible recorded outcome.
Treating a model assumption as a physical fact. Assigning equal probabilities to cards follows from the model’s shuffle assumption; it does not prove that a real shuffle is perfectly uniform.
Ignoring what was recorded. The outcome may be a card’s rank rather than its identity. Probabilities must match the specified recording rule.
Thinking probability describes a guaranteed short-run pattern. A probability assignment describes the model’s uncertainty about a result, not a promise about the order or balance of a small number of trials.
Forgetting to check normalization. In a finite model, the outcome probabilities must be nonnegative and add to one. If they do not, the assignment is not a probability model as specified.

Key takeaway: State the physical procedure, specify what is recorded, list the possible recorded outcomes, and identify the assumptions that assign probabilities to them. Those pieces together distinguish an experiment from its probability model.

Check Your Understanding

Answer each question by identifying the procedure, the recorded outcomes, or the model assumptions as appropriate.

A die is rolled once and the number showing is recorded. What is the physical experiment, and what counts as an outcome?
A finite model has five outcomes, all assigned the same probability. What is the probability of each outcome, and what condition justifies your answer?
A shuffled deck is used, but only whether a drawn card is red or black is recorded. Explain how the probabilities of red and black relate to probabilities assigned to individual cards.
A component inspection has a chance of misclassification. Which distinction in the worked inspection example would need to be represented differently?
Why does assigning probability \(1/2\) to heads in a coin model not imply that every two tosses contain one head?