The previous module ended with a question hanging in the air: the average satisfaction score of 8.1 comes from 640 responses out of Club Nova's roughly 380,000 members… how much can we trust that figure? To answer it we need a language for talking rigorously about uncertainty, and that language is probability. In this lesson we lay its foundations: what a random experiment is, how its possible outcomes are described (sample space and events), what operations we can perform on those events, and the three classical ways of attaching a number to the phrase "it is likely that". All of it with the NovaMarket data you already know: receipts, card payments, product categories and the satisfaction survey.

1. Random experiment and sample space
2. Events: elementary, compound, certain and impossible
3. Operations on events: union, intersection, complement and mutual exclusivity
4. The three approaches to probability: classical, frequentist and subjective
5. Kolmogorov's axioms (light version)
6. Probabilities from relative frequencies at NovaMarket

A random experiment is any process whose outcome cannot be predicted with certainty before it is carried out, even though we do know the set of possible outcomes. The opposite is a deterministic experiment: if Marta multiplies the average transaction value by yesterday's number of transactions, the result is always the same; there is no chance involved.

Random experiments are everywhere at NovaMarket:

• Pick a receipt at random from today's transactions and observe how it was paid (card, cash, other).
• Pick a Club Nova member at random and observe whether they answered the satisfaction survey.
• Observe how many hours the next online order takes to be delivered.
• Ask a customer at Madrid-Centro whether they would recommend the store.

The sample space, denoted \( \Omega \) (omega), is the set of all possible outcomes of the experiment. Each individual outcome is called an elementary event.

Notice that defining \( \Omega \) properly is the first step of any probabilistic analysis: if Marta asks "what is the probability that a receipt is paid by card?", we first have to agree that the possible outcomes are {card, cash, other} and not, say, {card, not card} — both choices are valid, but they change the calculations, and we must stay consistent.

An event is any subset of the sample space: a statement about the outcome that may or may not occur. Events are usually denoted with capital letters: \( A \), \( B \), \( C \)…

Take the experiment "satisfaction score of a randomly chosen response" (\( \Omega = {0,1,\dots,10} \)):

• Elementary event: consists of a single outcome. \( A = {10} \): "the score is exactly 10".
• Compound event: consists of several outcomes. \( B = {8, 9, 10} \): "the score is high (≥ 8)". \( C = {0,1,2,3,4,5} \): "the score is low (≤ 5)", the group of detractors we already identified in Module 2.
• Certain event: \( \Omega \) itself. "The score is between 0 and 10" always occurs.
• Impossible event: the empty set, \( \varnothing \). "The score is 12" can never occur.

We say an event occurs when the outcome of the experiment belongs to it. If the chosen response has a score of 9, then \( B \) has occurred (because \( 9 \in B \)) and so has the certain event, but neither \( A \) nor \( C \) has.

Events are sets, so they combine through set operations. This is the vocabulary we will use constantly in the next lesson.

And one important relationship between two events:

• \( A \) and \( B \) are mutually exclusive (or disjoint) when they cannot occur at the same time: \( A \cap B = \varnothing \).

We pick one of the survey's 640 responses at random. Define:

• \( T \) = "the response came through the in-store channel"
• \( O \) = "the response came through the online channel"
• \( S \) = "the score is ≥ 8" (satisfied)

Then:

• \( T \cup O \): the response came in-store or online (that is, through any channel other than the app).
• \( T \cap S \): the response came in-store and the score is ≥ 8. A perfectly possible event: channel and score are different attributes of the same outcome.
• \( S^{c} \): the score is ≤ 7.
• \( T \) and \( O \) are mutually exclusive: the same response cannot have arrived through two channels at once. \( T \) and \( S \), on the other hand, are not mutually exclusive.

Visually, a diagram helps fix the difference between mutually exclusive events (separate circles) and compatible ones (overlapping circles):

Two useful identities worth memorizing (they are De Morgan's laws, though the name matters less than the idea):

\[ (A \cup B)^{c} = A^{c} \cap B^{c} \qquad\qquad (A \cap B)^{c} = A^{c} \cup B^{c} \]

In words: "not (A or B)" means "neither A nor B"; and "not (A and B)" means "at least one of the two fails".

The probability of an event \( A \), written \( P(A) \), is a number between 0 and 1 measuring how likely it is to occur: 0 = impossible, 1 = certain, 0.5 = as likely as not. The question is where that number comes from. There are three classical answers, and all three are used in business.

When all elementary events are equally likely (by symmetry: dice, coins, prize draws):

\[ P(A) = \frac{\text{number of outcomes favorable to } A}{\text{number of possible outcomes}} \]

NovaMarket example. For Club Nova's monthly prize draw, one winner is picked completely at random from the 380,000 members. The probability that any particular member wins is \( 1/380{,}000 \approx 0.0000026 \). And if, say, 9,500 members are registered at the Cuenca store, the probability that the winner is from Cuenca is \( 9{,}500/380{,}000 = 0.025 \), or 2.5%: favorable outcomes (Cuenca members) over possible outcomes (all members), because the draw guarantees that everyone is equally likely.

The limitation is obvious: almost nothing in business is "equally likely by symmetry". The probability that a receipt is paid by card is not 1/3 just because there are three payment methods; the three outcomes do not carry the same weight. That is where the second approach comes in.

The probability of \( A \) is the value its relative frequency approaches when the experiment is repeated many times:

\[ P(A) \approx \frac{\text{number of times } A \text{ occurs}}{\text{number of repetitions}} \quad \text{(over many repetitions)} \]

This is the direct connection with Module 2: with enough data, the relative frequencies in a frequency table are estimates of probabilities. The fact that 61% of NovaMarket's receipts are paid by card — computed over millions of receipts — entitles us to say: "the probability that the next receipt is paid by card is approximately 0.61".

The theoretical guarantee that this works is called the law of large numbers: as the number of repetitions grows, the relative frequency settles down around the true probability. With 10 receipts we might observe 40% card payments by pure chance; with 100,000 receipts, the observed proportion will almost certainly sit right next to 61%.

Sometimes the experiment is not repeatable and we still need a probability. "What is the probability that opening store number 43 in Valencia will be profitable in its first year?" admits no frequencies: we will only open that store once. Subjective probability is an expert's degree of belief, expressed as a number between 0 and 1, based on their information and experience. Marta can say "I'd put it at 0.7" and that number is useful for deciding, as long as it is coherent (for instance, the probabilities of the possible scenarios must add up to 1).

Whichever approach you take, for a set of numbers to deserve the name "probability" it must satisfy three minimal rules, formulated by Andrei Kolmogorov in 1933. There is no need to memorize the full formalization; the three ideas are what count:

1. Non-negativity: \( P(A) \geq 0 \) for every event \( A \). Negative probabilities do not exist.
2. Normalization: \( P(\Omega) = 1 \). The certain event has probability 1.
3. Additivity: if \( A \) and \( B \) are mutually exclusive, \( P(A \cup B) = P(A) + P(B) \). The probabilities of events that cannot coincide add up.

From these three axioms follow consequences we use constantly:

• \( P(\varnothing) = 0 \): the impossible event has probability 0.
• \( P(A^{c}) = 1 - P(A) \): the complement rule, the most profitable of them all in practice.
• \( 0 \leq P(A) \leq 1 \): no probability exceeds 1.

A check with NovaMarket. The share of sales by category (fresh food 38%, pantry 27%, beverages 14%, household and personal care 12%, other 9%) works as a probability assignment for the experiment "category of a randomly chosen euro of sales": all shares are ≥ 0 and they add up to \( 0.38+0.27+0.14+0.12+0.09 = 1.00 \). ✔ And since "fresh food" and "beverages" are mutually exclusive (each euro is classified into a single category), \( P(\text{fresh food} \cup \text{beverages}) = 0.38 + 0.14 = 0.52 \).

Let's close the lesson by applying the frequentist approach to figures we already know from Module 2. We pick one of the 640 responses to the satisfaction survey at random:

• \( P(\text{score} \geq 8) = 0.712 \) (71.2% of the responses).
• \( P(\text{score} \leq 5) = 0.075 \) (the detractors).
• By the complement rule: \( P(\text{score} \geq 6) = 1 - 0.075 = 0.925 \).
• \( P(\text{channel} = \text{in-store}) = 448/640 = 0.70 \); \( P(\text{online}) = 128/640 = 0.20 \); \( P(\text{app}) = 64/640 = 0.10 \). Axiomatic check: \( 0.70+0.20+0.10 = 1 \). ✔

And with the receipts: if we pick a receipt at random, \( P(\text{card}) = 0.61 \), so \( P(\text{not card}) = 1 - 0.61 = 0.39 \).

One important nuance we will return to in Module 5: when the frequency comes from a sample (640 responses) rather than the whole population (380,000 members), the 0.712 is an estimate \( \hat{p} \) of the true probability \( \pi \), not the exact probability. The parameter/statistic distinction from Module 1 reappears here in a new disguise.

• Confusing "possible outcomes" with "equally likely outcomes". Having three payment methods does not give each one probability 1/3. Laplace's rule only applies when equal likelihood is guaranteed (prize draws, symmetric physical chance).
• Probabilities that don't add up to 1. If you assign probabilities to an exhaustive, mutually exclusive breakdown (channels, categories, payment methods) and they don't sum to 1, there is guaranteed to be an error somewhere. Use it as a routine check.
• Forgetting the complement. Computing \( P(\text{score} \geq 6) \) by adding seven probabilities is laborious and error-prone; \( 1 - P(\text{score} \leq 5) \) gives the same answer in one step.
• Reading probability as short-run certainty. \( P(\text{card}) = 0.61 \) does not mean that exactly 61 of the next 100 receipts will be paid by card; it means that in the long run the proportion will hover around 61%. Over short streaks, variability rules (we will quantify it in Module 4).
• Taking a small sample's relative frequency as an exact probability. The 64 members from the app channel are few: their average of 7.9 (and any proportion computed from them) carries far more uncertainty than the in-store channel's figures. Always note the n alongside the estimated probability.

Exercise 1. A NovaMarket receipt is chosen at random and its payment method observed, with \( \Omega = {\text{card}, \text{cash}, \text{other}} \), \( P(\text{card}) = 0.61 \) and \( P(\text{cash}) = 0.33 \). (a) Compute \( P(\text{other}) \). (b) Are "card" and "cash" mutually exclusive? (c) Compute \( P(\text{card} \cup \text{cash}) \) and \( P(\text{not card}) \).

Exercise 2. We pick a survey response at random (640 in total). Classify each event as elementary, compound, certain or impossible, and give its probability where the lesson's data allow: (a) "the score is between 0 and 10"; (b) "the score is ≥ 8"; (c) "the response came through the app"; (d) "the score is −2".

Exercise 3. Marta states: "The probability that the autumn campaign beats its target is 0.6; that it hits the target exactly, 0.3; and that it falls short, 0.2." (a) Which approach to probability is she using? (b) Is there anything wrong with her assignment? Justify your answer using the axioms.

Solution 1. (a) The three methods form an exhaustive, mutually exclusive breakdown, so their probabilities sum to 1: \( P(\text{other}) = 1 - 0.61 - 0.33 = 0.06 \). (b) Yes: a receipt is recorded with a single payment method, so \( \text{card} \cap \text{cash} = \varnothing \). (c) By additivity of mutually exclusive events, \( P(\text{card} \cup \text{cash}) = 0.61 + 0.33 = 0.94 \); and \( P(\text{not card}) = 1 - 0.61 = 0.39 \). Common mistake: answering \( P(\text{other}) = 0 \) "because it isn't given" — when a figure is missing, you deduce it, you don't ignore it.

Solution 2. (a) Certain event: \( P = 1 \). (b) Compound (it groups the scores 8, 9 and 10): \( P = 0.712 \). (c) Compound with respect to the full experiment (there are many app responses), with \( P = 64/640 = 0.10 \); it would be elementary only if the experiment observed nothing but the channel, with \( \Omega = {\text{in-store}, \text{online}, \text{app}} \) — the classification depends on how the sample space is defined, and either answer, well argued, is valid. (d) Impossible: \( P = 0 \). Common mistake: calling any event described in a single sentence "elementary"; what counts is how many outcomes of \( \Omega \) it contains.

Solution 3. (a) The subjective approach: this year's autumn campaign happens only once, there are no frequencies to consult. (b) Yes, there is a problem: the three scenarios are mutually exclusive and exhaustive, but \( 0.6 + 0.3 + 0.2 = 1.1 > 1 \), which violates the axioms (the certain event would have probability 1.1). Marta needs to revise her numbers — for instance, 0.5/0.3/0.2. Tip: subjective probability is legitimate, but it is obliged to be coherent; the axioms are the quality control for any assignment, wherever it comes from.

We now have the basic vocabulary of uncertainty: random experiments whose set of outcomes is the sample space \( \Omega \); events that are subsets of \( \Omega \) and combine through union, intersection and complement; and three ways of assigning probabilities — Laplace when outcomes are equally likely, relative frequencies when data are plentiful (that is how we justified the 0.61 for card payments and the 0.712 for satisfied customers), and expert judgment when we have neither — all subject to Kolmogorov's axioms, which guarantee coherence.

But the interesting business questions are almost never about a single event in isolation: they are about combinations. What is the probability that a response comes from the app and belongs to an unhappy customer? If we know an order arrived late, does that change the probability that a particular carrier delivered it? To answer, we need the Probability Rules: the addition rule, conditional probability, the multiplication rule and Bayes' theorem.