Weeks 1–2 described what already happened. From this week on, the pizzeria has to plan on chances: how many Margheritas will a Friday bring, what is an insurance premium worth, how likely is a sell-out? This week builds the language for that — probability rules, random variables, their mean and standard deviation, and the first workhorse distribution, the binomial. This is the textbook's chapter 4, opened with the rules from chapter 3 it relies on.
The session is three hours in two parts. Part 1 (~1 h) is the lesson: your instructor presents the deck and you follow the same numbered sections here — every slide names its section. Part 2 (~2 h) is ten exercises taken straight from the textbook, worked together; solutions open as your instructor reveals them.
On a Friday the pizzeria expects 140 customers. Each of them, independently, orders a Margherita with probability 0.8. The kitchen prepares 120 dough balls for Margheritas in the morning, and dough cannot be made during service. What is the chance we run out?
The number of Margheritas ordered on Friday is not known in advance. It is a random variable: a number that depends on chance. To say how likely 121 or more is, you need its whole probability distribution — every possible value and its probability. Section 3.6 writes that distribution down, and the answer comes back at the end of it.
Description told you what happened. Probability tells you what to prepare for.
Every order is classified by what was ordered and where it went. When every order is equally likely to be picked, a probability is simply a count divided by 400. Choose an event A (a row) and an event B (a column) and read off every rule of chapter 3.
Mutually exclusive and independent are not the same thing — they are nearly opposites. If A and B cannot happen together, learning that B happened tells you A did not.
"How many Margheritas will Friday bring?" has an answer only after Friday. Before it, the count is a random variable — written with a capital letter, X, while a particular value it takes is a lower-case x.
Count → discrete. Measure → continuous. This week is discrete; Week 4 is continuous.
Two traps. A count can be huge and still be discrete — the number of kernels in a bag of popcorn. And money is treated as continuous when it is an amount (the average spent on electricity), because it can take essentially any value in a range.
For it to be a valid distribution: every P(x) is between 0 and 1, and the P(x) add up to exactly 1. Nothing else is required — the values x need not be whole numbers.
The mean µ is a weighted average: each value weighted by how often it happens. The standard deviation σ is the same idea as in Week 2 — the typical distance from the mean — with probabilities in place of 1/n.
µ = 2.6 Margheritas an hour. Over a five-hour Friday service that is 13 — the number the morning prep list is built on.
When X is a net gain — winnings minus the price of playing — its mean has a special name, the expected value E(X). It is not a value you expect to see on any single play; it is the long-run average. Insurers, lotteries and casinos set prices by it.
The pizzeria's oven costs CHF 12,000 to replace. An insurer estimates the chance it breaks down within a year at 3%, and sells a one-year policy for a premium C. For the insurer, X takes two values:
| outcome | no breakdown | breakdown |
|---|---|---|
| x (insurer's net gain) | C | C − 12,000 |
| P(x) | 0.97 | 0.03 |
E(X) = C − 360. Below CHF 360 the insurer loses money on average; everything above it pays for staff, risk and profit.
A lottery ticket costs $1 and the prize is $1,000: the winner's net gain is $999, not $1,000, and everybody else's is −$1. Forgetting to subtract the stake is the most common mistake in Exercises 5 and 6 of Part 2 — and it moves the answer by exactly the price of the ticket.
Read the formula left to right: C(n, k) counts the orders in which k successes can fall among n trials; pᵏ is the chance of k successes; qⁿ⁻ᵏ the chance that the other n − k fail. Ten tables, each ordering dessert with p = 0.3: P(exactly 3) = C(10, 3) · 0.3³ · 0.7⁷ = 120 · 0.027 · 0.0824 = 0.2668.
X = number of Margheritas ordered. Fixed n = 140, two outcomes, the same p = 0.8 for everybody, customers deciding independently: X is binomial. We sell out when X ≥ 121, i.e. X > 120.
This week's exercises are not invented: each one is reproduced exactly as it appears at the end of a section of the textbook, and each is labelled with its section and number, so you can find it in the book. Your instructor puts each exercise on the slide; you work it here, by hand, with a calculator. Check marks your number immediately, Hint nudges without giving it away, and the Solution button opens when the room does that exercise together.
About 110 minutes for all ten. Give probabilities to four decimals, as the book does. Parts of an exercise that ask you to explain are answered in the worked solution rather than in a box. Your answers are saved and reach your instructor's dashboard. Nothing here is graded.
One attempt per question, every answer explained. Your score syncs to your instructor's dashboard as a self-check but does not count toward your grade. Have a calculator to hand.
In Week 4 the random variable becomes continuous — dough weight, waiting time — and the bars of this week melt into a curve: the normal distribution. It also answers the sell-out question for a 140-customer Friday in a line, because a binomial with large n looks normal. Before that, the Week 3 homework: thirteen more exercises from the book, by hand.
140 customers, each ordering a Margherita with p = 0.8. Every dough ball prepared and not used is thrown away at closing; every Margherita you cannot make is a disappointed customer. Move the slider and see the chance of selling out against the expected number of dough balls wasted.
There is no risk-free number. Statistics does not choose for you — it tells you exactly what each choice costs.
📎 Everything saves automatically and reaches your instructor's dashboard. This is the third instalment of the individual project.
Applied Statistics · Week 3 · Discrete Random Variables · Jan Erik Meidell · HEG Genève
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