Module 3 closed by planting a decisive idea: many chance phenomena follow such well-worn patterns that there is no need to model them from scratch, because ready-made templates with names of their own already exist. This lesson introduces the first one, and the most natural one for NovaMarket: the binomial distribution, the model for counting successes in a fixed number of independent trials. How many of the next 10 receipts will be paid by card? How many of this afternoon's 20 orders will arrive past the deadline? How many of the 380,000 Club Nova coupons will be redeemed? All three questions are, deep down, the same question, and a single formula answers them all. You will learn where that formula comes from, how to apply it step by step, what mean and what spread the model has, and — every bit as important — how to recognize when a situation is not binomial.

1. The basic building block: the Bernoulli trial
2. The four conditions of the binomial model
3. The binomial formula and the binomial coefficient
4. A complete example: card-paid receipts in Madrid-Centro
5. Mean and variance of a binomial
6. A business example: deliveries outside the SLA
7. The shape of the distribution as \( p \) changes
8. When it is NOT binomial

A Bernoulli trial is the simplest possible random experiment: one with only two outcomes, which by convention we call "success" and "failure". In the previous lesson we already worked with variables of this kind without naming them: \( Y = 1 \) if the receipt is paid by card, \( Y = 0 \) if not.

• "Success" is a technical label, not a value judgment: if we are studying late deliveries, the "success" may well be "arriving late". Success = the event we are counting.
• The probability of success is denoted \( p \); the probability of failure, \( q = 1 - p \).

A Bernoulli variable \( Y \) has a minimal distribution table:

Applying the expected value and variance formulas from the previous lesson: \( E(Y) = 0 \cdot (1-p) + 1 \cdot p = p \) and \( \operatorname{Var}(Y) = (0-p)^2(1-p) + (1-p)^2 p = p(1-p) \). Both are worth memorizing: they will reappear multiplied by \( n \) in a moment.

Bernoulli trials at NovaMarket, with figures we already know:

The binomial model arises from repeating a Bernoulli trial and counting the successes. We say that \( X \) follows a binomial distribution with parameters \( n \) and \( p \) — written \( X \sim B(n, p) \) — when these four conditions hold:

1. Fixed number of trials: \( n \) is decided in advance (the next 10 receipts, today's 20 orders).
2. Two outcomes per trial: each trial is a Bernoulli trial (success/failure).
3. Constant probability: \( p \) is the same in every trial.
4. Independence: the outcome of one trial does not influence the others (recall the formal definition of independence from lesson 03-02).

If all four hold, the variable \( X \) = "number of successes in the \( n \) trials" takes values in \( {0, 1, 2, \dots, n} \) and its pmf has a closed-form formula: we no longer need to estimate the distribution table value by value, as we did with the reusable bags, because the template generates the whole thing from just two numbers, \( n \) and \( p \). That is exactly the advantage of distribution models we announced at the close of Module 3.

\[ P(X = k) = \binom{n}{k} , p^k , (1-p)^{n-k} \qquad k = 0, 1, \dots, n \]

The formula has three factors, and each one answers a question:

• \( p^k \): the probability that one specific sequence has successes in \( k \) given positions. By the product rule for independent events (Module 3), the probabilities multiply.
• \( (1-p)^{n-k} \): the probability of failure in the other \( n-k \) positions of that same sequence.
• \( \binom{n}{k} \): the binomial coefficient ("\( n \) choose \( k \)"), which counts how many distinct sequences contain exactly \( k \) successes. Since all those sequences are mutually exclusive and equally likely, the addition rule tells us that multiplying is all it takes.

The binomial coefficient is computed as:

\[ \binom{n}{k} = \frac{n!}{k! , (n-k)!} \]

where \( n! = n \times (n-1) \times \dots \times 2 \times 1 \) (and by convention \( 0! = 1 \)). A mini-example to pin the idea down: in how many ways can 2 out of 3 receipts be card payments? The possible sequences are CCx, CxC and xCC: three. And indeed:

\[ \binom{3}{2} = \frac{3!}{2! \cdot 1!} = \frac{6}{2 \times 1} = 3 \]

Two properties that save calculations: \( \binom{n}{0} = \binom{n}{n} = 1 \) (there is only one way to fail everything or to succeed at everything) and \( \binom{n}{k} = \binom{n}{n-k} \) (choosing which \( k \) positions are successes is the same as choosing which \( n-k \) are failures).

Marta requests an analysis to size the card terminal fleet: of the next 10 receipts at a Madrid-Centro checkout, how many will be paid by card? We know that 61% of receipts are paid by card, and it is reasonable to assume that customers decide independently of one another. Therefore:

\[ X = \text{"number of card-paid receipts out of the next 10"} \sim B(10;\ 0.61) \]

Step by step for \( P(X = 6) \):

1. Binomial coefficient: \( \binom{10}{6} = \frac{10!}{6! \cdot 4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5{,}040}{24} = 210 \).
2. Success power: \( 0.61^6 \approx 0.05152 \).
3. Failure power: \( 0.39^4 \approx 0.02313 \).
4. Product: \( P(X = 6) = 210 \times 0.05152 \times 0.02313 \approx 0.2503 \).

There is a 25% probability that exactly 6 of the 10 will pay by card. Repeating the calculation for each \( k \) yields the complete distribution:

(The sum is 1, as every pmf requires.) With the table, any question is answered by adding, just as in the previous lesson:

• "What if at least 8 pay by card?": \( P(X \geq 8) = 0.1312 + 0.0456 + 0.0071 = 0.1839 \).
• "And 4 or fewer?": \( P(X \leq 4) = 0.0001 + 0.0013 + 0.0090 + 0.0374 + 0.1023 = 0.1501 \).

Note a trick we will use constantly: for "at least one", the complement is almost always the shorter route. \( P(X \geq 1) = 1 - P(X = 0) = 1 - 0.39^{10} \approx 1 - 0.0001 = 0.9999 \).

We could compute \( E(X) \) from the table above (summing \( k \cdot P(X=k) \)), but the template comes with the formulas built in. Since \( X \) is the sum of \( n \) independent Bernoulli variables, its expected value is \( n \) times that of one Bernoulli trial, and likewise its variance:

\[ E(X) = np \qquad \operatorname{Var}(X) = np(1-p) \qquad \sigma = \sqrt{np(1-p)} \]

For the 10 receipts: \( E(X) = 10 \times 0.61 = 6.1 \) card-paid receipts, with \( \sigma = \sqrt{10 \times 0.61 \times 0.39} = \sqrt{2.379} \approx 1.54 \). The analyst's reading: "we expect about 6 out of 10, with a typical fluctuation of ±1.5". Look at the table and check that almost all the probability is concentrated between \( 6.1 - 2 \times 1.54 \approx 3 \) and \( 6.1 + 2 \times 1.54 \approx 9 \): indeed, \( P(3 \leq X \leq 9) \approx 0.98 \).

A useful detail of the variance formula: for fixed \( n \), \( p(1-p) \) is largest when \( p = 0.5 \). Uncertainty about the count is greatest when success and failure are evenly matched, and smallest when \( p \) approaches 0 or 1 (with \( p = 0.005 \), as with fraud, almost every count will be 0).

Recall the e-commerce SLA: "90% of orders in under 48 h". In Module 3 we read off the CDF of delivery times that \( F(48) = 0.899 \), so the probability that a given order arrives past the deadline is \( p \approx 0.10 \).

The e-commerce team will dispatch 20 orders this afternoon to different customers, split across carriers as usual. Marta asks: "what is the probability that 3 or more arrive late?" We model \( X \sim B(20;\ 0.10) \) and use the complement:

\[ P(X \geq 3) = 1 - P(X=0) - P(X=1) - P(X=2) \]

1. \( P(X=0) = \binom{20}{0} , 0.1^0 , 0.9^{20} = 0.9^{20} \approx 0.1216 \)
2. \( P(X=1) = \binom{20}{1} , 0.1^1 , 0.9^{19} = 20 \times 0.1 \times 0.1351 \approx 0.2702 \)
3. \( P(X=2) = \binom{20}{2} , 0.1^2 , 0.9^{18} = 190 \times 0.01 \times 0.1501 \approx 0.2852 \)
4. \( P(X \geq 3) = 1 - (0.1216 + 0.2702 + 0.2852) = 1 - 0.6770 = 0.3230 \)

Nearly one day in three, the afternoon will end with 3 or more late deliveries — even though the expected count is only \( E(X) = 20 \times 0.10 = 2 \). This result is invaluable for managing expectations: if Marta got a complaint every time an afternoon produced 3 delays, the binomial tells her that this will happen frequently even when the process is working exactly as designed. Telling normal fluctuation apart from genuine service deterioration is, precisely, the heart of Module 5.

An honest caveat about the model: here we treat \( p = 0.10 \) as constant, but we know from Module 3 that lateness depends on the carrier (LogiExpress 8%, RápidoSur 15%). If the 20 orders are split in the usual proportions (70/30), the overall \( p \) of 10% is a valid weighted average and the model works well; if on a given day everything went out with RápidoSur, we would have to use \( B(20;\ 0.15) \). The model is a template: it must be filled in with the right \( p \) for the context.

The binomial does not always have the same silhouette. Using the ideas about shape from Module 2 (symmetry and skewness of a histogram):

And one observation we leave planted: for large \( n \), even when \( p \) is not 0.5, the distribution becomes increasingly symmetric and bell-shaped. Check it mentally with the coupons: \( B(380{,}000;\ 0.18) \) has \( E(X) = 68{,}400 \) redemptions with \( \sigma = \sqrt{380{,}000 \times 0.18 \times 0.82} \approx 237 \); nobody would compute \( \binom{380{,}000}{68{,}400} \) by hand. That emerging bell has a name of its own — we will meet it in the next lesson — and the reason the approximation works is explained by the central limit theorem.

Recognizing when the template does not fit is as important as knowing how to use it. The three typical failure modes attack conditions 3 and 4:

• Sampling without replacement from a small population (independence and constant \( p \) both fail). If a crate of 12 units of fruit contains 3 spoiled ones and I draw 4 without putting them back, the probability of drawing a defective one changes with each draw (3/12, then perhaps 2/11…). The correct model is the hypergeometric distribution, which we will meet in 04-03. Rule of thumb: if the sample is less than 10% of the population, the binomial is an excellent approximation — which is why treating 10 receipts out of the day's thousands as independent is legitimate.
• Probability that is not constant across trials. The 61% card-payment rate is an annual average; a Friday evening in Valencia-Ruzafa may run at 70% and a Tuesday morning in Cuenca at 50%. Counting "card-paid receipts over a whole week, mixing time slots" with a single \( p \) can bias the results if the slots carry different weight on different days.
• Dependence between trials. Customers do not always decide alone: if the store's card terminal goes down, all subsequent receipts switch to cash at once; a family shopping together generates correlated receipts; in fraud, attacks arrive in coordinated bursts. When the outcome of one trial drags the others along, the binomial underestimates the probability of extreme outcomes.

And a reminder about scope: the binomial counts successes in a fixed number of trials. If the question is "how many attempts until the first success?" or "how many events per hour?", the template is a different one (geometric and Poisson, both in the next lesson on other distributions).

• Forgetting the binomial coefficient. \( p^k (1-p)^{n-k} \) is the probability of one specific sequence; without the \( \binom{n}{k} \) factor the result comes out absurdly small. If your probability of "6 of 10 by card" comes out as 0.0012 instead of 0.25, you almost certainly left out the 210.
• Confusing "at least \( k \)" with "exactly \( k \)". \( P(X \geq 3) \neq P(X = 3) \). Always translate the business phrase into an inequality before calculating, and consider the complement shortcut: for "at least one", \( 1 - P(X=0) \).
• Swapping success and failure mid-calculation. If you defined success = "arrives late" with \( p = 0.10 \), then \( 0.9^{20} \) is the probability that none arrive late, not that all arrive late. Write down the definition of success at the start and do not change it.
• Applying the binomial to counts without a fixed \( n \). "Customers per hour at the checkout" has no predetermined number of trials: it is not binomial (it is Poisson, next lesson).
• Tip: always sanity-check the order of magnitude with \( E(X) = np \) before reporting a result. If you expect 2 delays out of 20 orders and your calculation says the most likely count is 9, something is wrong.
• Tip: in reports for management, pair the expected value with \( \sigma \). "We expect 2 delays, and up to 4–5 falls within normal variation" prevents unwarranted alarms.

Exercise 1. 45% of NovaMarket receipts include fresh produce. For a fruit promotion test, the next 8 receipts at Sevilla-Nervión will be observed. (a) Justify that the binomial model applies and identify \( n \) and \( p \). (b) Compute the probability that exactly 4 receipts include fresh produce. (c) Compute \( E(X) \) and \( \sigma \).

Exercise 2. The annual Club Nova churn rate is 3%. An account manager in Cuenca has a portfolio of 10 "premium" members he looks after personally, and boasts that he has not lost a single one this year. (a) If his members behaved like the average member (independent cancellations, \( p = 0.03 \)), what was the probability of ending the year with no cancellations? (b) And of suffering at least one? (c) In light of (a), does his result prove he manages better than average?

Exercise 3. RápidoSur delivers 15% of its shipments late. Tomorrow it has 12 orders assigned. (a) Compute the probability that at most 1 arrives late. (b) Compute the expected number of delays and its standard deviation. (c) Marta proposes measuring RápidoSur by counting the delays on its next 12 shipments to the same office building in the same van. Which condition of the binomial model suffers?

Solution 1. (a) There are \( n = 8 \) trials fixed in advance, each receipt either includes fresh produce or not (two outcomes), \( p = 0.45 \) is constant and receipts from different customers are reasonably independent: \( X \sim B(8;\ 0.45) \). (b) \( \binom{8}{4} = \frac{8!}{4! , 4!} = 70 \); \( 0.45^4 \approx 0.04101 \); \( 0.55^4 \approx 0.09151 \); \( P(X=4) = 70 \times 0.04101 \times 0.09151 \approx 0.2627 \). (c) \( E(X) = 8 \times 0.45 = 3.6 \) receipts; \( \sigma = \sqrt{8 \times 0.45 \times 0.55} = \sqrt{1.98} \approx 1.41 \). Common mistake: using \( 0.45^4 \times 0.55^4 \) without the 70 — that would give 0.0038, the probability of one single specific sequence.

Solution 2. (a) \( P(X = 0) = 0.97^{10} \approx 0.7374 \). (b) \( P(X \geq 1) = 1 - 0.7374 = 0.2626 \). (c) No: an average manager would end the year with no cancellations almost 3 times out of 4. The result is entirely compatible with not being better than average. Common mistake: interpreting a good outcome that is nevertheless likely under pure chance as proof of merit — the same fallacy as the base-rate fallacy from the fraud case. How to decide formally whether a result "proves" something is the subject of hypothesis testing in Module 5.

Solution 3. (a) \( X \sim B(12;\ 0.15) \). \( P(X=0) = 0.85^{12} \approx 0.1422 \); \( P(X=1) = 12 \times 0.15 \times 0.85^{11} \approx 12 \times 0.15 \times 0.1673 \approx 0.3012 \); \( P(X \leq 1) \approx 0.4435 \). (b) \( E(X) = 12 \times 0.15 = 1.8 \); \( \sigma = \sqrt{12 \times 0.15 \times 0.85} = \sqrt{1.53} \approx 1.24 \). (c) Independence: if the van gets a flat tire or the driver runs late, all 12 orders arrive late at once. With dependent trials, the binomial underestimates the extreme scenarios (0 delays or 12 delays would be far more likely than the model predicts).

The binomial is your first template: when you count successes in \( n \) independent trials with constant \( p \), you no longer need to build the distribution table by hand — the formula \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \) generates the whole thing, and \( E(X) = np \) and \( \sigma = \sqrt{np(1-p)} \) summarize where it is centered and how much it fluctuates. You have also seen its limits: without independence, or with a varying \( p \), the template misleads. And a curious phenomenon has come into view: as \( n \) grows, the binomial's stepped silhouette smooths out until it looks like a symmetric bell. That bell is no accident: it is the most important distribution in all of statistics, the one you already sensed when looking at the daily sales histogram in Module 2. We give it a name, a formula and a table in the next lesson: the Normal Distribution.