Week 2 — Sample spaces, events & probability rules

Outcomes, events as sets, and the complement / addition rules

The week question

Last week we agreed on what a probability means: a number between 0 and 1 that grades how strongly we expect a thing to happen. This week the question is sharper and more practical:

Once we have written down every way a situation could turn out, how do we describe the particular things we care about — and how do we compute their probabilities by combining the things we already know?

The answer is to treat outcomes as a list, events as subsets of that list, and probability as a rule that respects how those subsets fit together. Two of those rules — the complement rule and the addition rule — do most of the everyday work, and we build them carefully here.

Why this matters

Almost every probability you will ever compute is really a question about a set of outcomes. “Will the shuttle be late?” is a question about which morning-outcomes count as late. “Will it rain or the shuttle run late?” is a question about combining two such sets. If you can name the outcomes, draw the events as regions, and keep track of where they overlap, then the arithmetic follows almost mechanically — and you stop guessing.

The payoff is also conceptual. Conditional probability (Week 3), independence (Week 4), and Bayes’ rule (Week 5) are all built on top of this set-based picture. The addition rule you learn this week is the first of a small family of “combining” rules; getting it exactly right now — especially the correction for double-counting — saves you from a recurring error later. This is the grammar of the language; the rest of the course is sentences.

Learning goals

By the end of the week you should be able to:

Write down a sample space \(\Omega\) for a small situation, listing or describing every outcome.
Express an event as a subset of \(\Omega\) and read its probability as the chance that the realized outcome lands in that subset.
Use the complement rule \(P(A^c) = 1 - P(A)\) to compute the probability of “not \(A\).”
Form unions \(A \cup B\) and intersections \(A \cap B\), and say in words what each one means.
Apply the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), and recognize when the \(P(A \cap B)\) correction vanishes because the events are mutually exclusive.
Use the equally-likely model \(P(A) = |A| / |\Omega|\) when — and only when — every outcome is genuinely equally likely.

Core vocabulary

Outcome. A single, complete way the situation could turn out. Exactly one outcome happens.
Sample space \(\Omega\). The set of all possible outcomes. Writing it down is usually the first and most clarifying step.
Event. Any subset \(A \subseteq \Omega\) — a collection of outcomes we have grouped together because we care about them as a unit. The event “happens” when the realized outcome is one of its members.
Complement \(A^c\). The event that \(A\) does not happen: every outcome in \(\Omega\) that is not in \(A\). (You may also see \(\bar{A}\) or \(A'\) elsewhere; this course writes \(A^c\).)
Union \(A \cup B\). The event “\(A\) or \(B\) (or both)” — outcomes in \(A\), in \(B\), or in both.
Intersection \(A \cap B\). The event “\(A\) and \(B\)” — outcomes in both \(A\) and \(B\).
Mutually exclusive (disjoint). Two events with no outcome in common: \(A \cap B = \varnothing\). They cannot both happen on the same trial.
Equally-likely model. A modeling assumption that every outcome in a finite \(\Omega\) carries the same probability \(1/|\Omega|\).

Concept development

Outcomes and the sample space \(\Omega\)

Start by asking: what are the distinct, complete results this situation could produce? Each such result is an outcome, and the collection of all of them is the sample space, written \(\Omega\). The discipline is that the list must be exhaustive (every possibility is in it) and the outcomes mutually exclusive (exactly one of them occurs on a given trial). Get those two properties right and everything downstream is well defined.

For a single fair coin, \(\Omega = \{H, T\}\). For one roll of a six-sided die, \(\Omega = \{1, 2, 3, 4, 5, 6\}\). The art is in choosing a sample space that is detailed enough to answer the questions you care about but no more detailed than that. If you only care whether a die shows an even number, you could in principle use \(\Omega = \{\text{even}, \text{odd}\}\) — but a finer space is often safer, because it keeps the equally-likely assumption honest, as we will see.

When a situation has two parts, the natural sample space is the set of ordered pairs of their results. If the first part has outcomes in a set \(S_1\) and the second has outcomes in \(S_2\), then

\[ \Omega = S_1 \times S_2 = \{(s_1, s_2) : s_1 \in S_1,\ s_2 \in S_2\}. \]

This “product” construction is exactly what we use for the commuter’s-morning slice below, where one part is the weather and the other is whether the shuttle runs on time.

Events as subsets, and the complement rule

An event is just a subset \(A \subseteq \Omega\). We bundle outcomes together and give the bundle a name because we care about it as a whole: “the die shows an even number” is the event \(A = \{2, 4, 6\}\). The probability of an event is the chance that the realized outcome is one of its members. Because every outcome carries some probability and the probabilities over all of \(\Omega\) sum to 1, two boundary facts hold for any model:

\[ P(\Omega) = 1, \qquad P(\varnothing) = 0. \]

The complement of \(A\), written \(A^c\), is the event that \(A\) does not happen — all the outcomes left over when you remove \(A\) from \(\Omega\). Since each outcome is either in \(A\) or in \(A^c\) but never both, their probabilities must account for the whole sample space:

\[ P(A) + P(A^c) = 1 \qquad \Longrightarrow \qquad P(A^c) = 1 - P(A). \]

This is the complement rule, and it is the single most useful shortcut in the course. Whenever the event you want is awkward but its opposite is easy, compute the opposite and subtract. “At least one” problems are the classic case: “at least one head in ten flips” is messy head-on, but its complement, “no heads at all,” is a single clean outcome.

Unions, intersections, and the addition rule

Most interesting events are built from simpler ones with the words or and and. In set language:

Union \(A \cup B\) is the event “\(A\) or \(B\),” meaning at least one of them — outcomes in \(A\), in \(B\), or in both.
Intersection \(A \cap B\) is the event “\(A\) and \(B\),” meaning both at once — outcomes that belong to each.

To get \(P(A \cup B)\) you might be tempted to add \(P(A)\) and \(P(B)\). But if the two events overlap, every outcome in the overlap \(A \cap B\) has been counted twice — once inside \(A\), once inside \(B\). The fix is to subtract that overlap back out exactly once. This is the addition rule (also called inclusion–exclusion for two events):

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]

Picture two overlapping circles inside a rectangle (the rectangle is \(\Omega\)). The union is the total shaded area; adding the two circles’ areas counts the lens-shaped overlap twice, so you subtract it once to get the true total. Two special cases are worth holding onto. When \(A\) and \(B\) are mutually exclusive — no shared outcome, so \(A \cap B = \varnothing\) and \(P(A \cap B) = 0\) — the rule collapses to the simpler

\[ P(A \cup B) = P(A) + P(B) \qquad \text{(mutually exclusive only)}. \]

And \(A\) and \(A^c\) are always mutually exclusive with union \(\Omega\), which is just the complement rule in disguise: \(P(A) + P(A^c) = P(\Omega) = 1\).

The equally-likely model \(P(A) = |A| / |\Omega|\)

When the sample space is finite and we have a good reason to believe every outcome is equally likely — a fair coin, a fair die, a well-shuffled deck — probability reduces to counting. If \(|\Omega|\) is the number of outcomes in the whole space and \(|A|\) is the number in the event \(A\), then

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

This is the classical or equally-likely model. It is powerful and it is the engine behind the counting techniques in Week 6 — but it carries a load-bearing assumption. The formula is only valid when the outcomes really are equally likely. A loaded die, or a sample space that lumps unequal possibilities together, breaks it. The commuter’s-morning case below is a deliberate example of a space that is not equally likely: rainy and dry mornings do not occur with equal frequency, so we cannot just count cells — we have to carry the actual probabilities. Knowing when the counting shortcut applies is as important as knowing the shortcut.

Worked examples

Worked example — the commuter’s morning, as a 2×2 sample space

Synthetic scenario; seed set for any simulation. Numbers are fixed for the whole course.

Maya’s morning has two moving parts: the weather and the shuttle. Let the weather outcome be either \(r\) (rain) or \(\bar r\) (no rain), and the shuttle outcome be either \(\text{ot}\) (on time) or \(\text{lt}\) (late). The natural sample space is the product of the two — a \(2 \times 2\) grid of four joint outcomes:

\[ \Omega = \{\text{on time}, \text{late}\} \times \{\text{rain}, \text{no rain}\} = \{(\text{ot}, r),\ (\text{ot}, \bar r),\ (\text{lt}, r),\ (\text{lt}, \bar r)\}. \]

Symbolically. Define the events we care about as subsets of \(\Omega\):

\(L = \{(\text{lt}, r),\ (\text{lt}, \bar r)\}\) — the shuttle is late (either weather).
\(R = \{(\text{ot}, r),\ (\text{lt}, r)\}\) — it is raining (either shuttle result).

The on-time event is the complement of late, \(L^c\), so the complement rule gives \(P(L) = 1 - P(L^c)\) — that is, \(P(\text{late}) = 1 - P(\text{on time})\).

Numerically. From the course case (Week 1), the marginal chance the shuttle is on time is \(P(\text{on time}) = 0.81\). The four joint-cell probabilities, which we will derive properly in Week 3, are fixed as:

weather \ shuttle	on time	late	row total
rain	\(0.18\)	\(0.12\)	\(0.30\)
no rain	\(0.63\)	\(0.07\)	\(0.70\)
column total	\(0.81\)	\(0.19\)	\(1.00\)

Notice the four cells sum to \(1\) — they are a valid probability over \(\Omega\) — but they are not all equal, so this is not an equally-likely model; we read probabilities off the table, not by counting cells. Now the complement rule does its job cleanly:

\[ P(\text{late}) = P(L) = 1 - P(\text{on time}) = 1 - 0.81 = 0.19. \]

That matches the \(P(\text{late}) = 0.19\) we locked in last week — arrived at here purely from the complement rule, with no new information.

An addition-rule example on the same grid. What is the chance the morning is “rough” in the sense that it is raining or the shuttle is late (or both)? That is the union \(R \cup L\). Reading the marginals and the overlap straight from the table:

\[ P(R) = 0.30, \qquad P(L) = 0.19, \qquad P(R \cap L) = 0.12, \]

where \(P(R \cap L) = 0.12\) is the single “rain and late” cell. The addition rule gives:

\[ P(R \cup L) = P(R) + P(L) - P(R \cap L) = 0.30 + 0.19 - 0.12 = 0.37. \]

Had we forgotten the overlap and simply added \(0.30 + 0.19 = 0.49\), we would have double-counted the rainy-and-late mornings and overstated the risk by \(0.12\). The subtraction is the whole point of the rule. Note too that \(R\) and \(L\) are not mutually exclusive — they share the \((\text{lt}, r)\) outcome — which is exactly why the correction term is nonzero here.

Worked example — a transfer: drawing a single card

Synthetic scenario in a fresh context; here the equally-likely model genuinely applies.

Now move to a standard, well-shuffled \(52\)-card deck and draw one card. Because a fair shuffle makes every card equally likely, the equally-likely model is justified and we may count: \(|\Omega| = 52\).

Symbolically. Let \(H\) be the event “the card is a heart” and \(F\) be the event “the card is a face card” (Jack, Queen, or King). Then \(H \cup F\) is “a heart or a face card,” and \(H \cap F\) is “a heart and a face card” — the face cards that are also hearts. The addition rule reads \(P(H \cup F) = P(H) + P(F) - P(H \cap F)\), with each piece computed by counting.

Numerically. Count the favorable outcomes:

Hearts: \(|H| = 13\), so \(P(H) = 13/52\).
Face cards: \(3\) in each of \(4\) suits, so \(|F| = 12\) and \(P(F) = 12/52\).
Hearts that are also face cards (Jack, Queen, King of hearts): \(|H \cap F| = 3\), so \(P(H \cap F) = 3/52\).

Apply the addition rule:

\[ P(H \cup F) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} \approx 0.423. \]

And the complement rule answers “not a heart” in one line:

\[ P(H^c) = 1 - P(H) = 1 - \frac{13}{52} = \frac{39}{52} = \frac{3}{4} = 0.75. \]

Compare the two worked cases: the card draw is equally likely, so we counted with \(P(A) = |A|/|\Omega|\); the commuter’s morning is not equally likely, so we read the probabilities off the table. Same two rules, two different ways of getting the inputs.

A common mistake

The signature Week 2 error is forgetting the overlap in the addition rule — writing \(P(A \cup B) = P(A) + P(B)\) when \(A\) and \(B\) can both happen. That shortcut is valid only for mutually exclusive events; whenever the two events can co-occur, you are double-counting the intersection and your answer is too large. In the card example, “heart or face card” is \(0.423\), not \(13/52 + 12/52 = 25/52 \approx 0.481\); the missing \(-3/52\) is the three cards counted twice. The fix is a habit: before adding, ask “can both events happen at once?” If yes, find \(P(A \cap B)\) and subtract it. If genuinely no, then \(P(A \cap B) = 0\) and the correction is harmless. A close cousin of this mistake is leaning on \(P(A) = |A|/|\Omega|\) when the outcomes are not equally likely — which is why we never count cells in the commuter’s-morning grid.

Low-stakes self-checks (ungraded)

Work these for yourself — they are practice only, with no points and nothing to submit. Sketch a Venn diagram or a small table before reaching for a formula.

Write out the sample space \(\Omega\) for flipping a fair coin twice as ordered pairs. How many outcomes are there? List the event “exactly one head” as a subset.
For Maya’s \(2 \times 2\) grid, use the complement rule to find \(P(\text{no rain})\) from \(P(\text{rain}) = 0.30\). Then identify which single cell is \(P(\text{on time} \cap \text{rain})\).
On the grid, compute \(P(\text{on time} \cup \text{no rain})\) with the addition rule. Which cell is the overlap you must subtract?
Draw one card from a standard deck. Find \(P(\text{red} \cup \text{king})\) using the addition rule, being careful about the two red kings. Then find \(P(\text{not a king})\) by the complement rule.
Give an example of two events from the card draw that are mutually exclusive, and explain why their addition rule has no subtraction term.

Reading and source pointer

This week corresponds to Grinstead & Snell, Chapter 1 — Discrete Probability Distributions, which introduces sample spaces, events, and the basic rules for combining their probabilities; the equally-likely model and the set operations are also covered cleanly in the MIT OCW 18.05 reading on counting and sets. Read those for additional examples and a second voice on the same ideas. These notes are the course’s own synthesis, grounded in but not copied from the sources. All scenario data are synthetic, with seeds set in any accompanying simulation so results are reproducible.

The companion Lab 2 — Monte Carlo basics turns these rules into code: you will simulate the commuter’s-morning grid many times and watch the long-run frequency of an event approach the probability the addition and complement rules predict — your first concrete look at the link between probability and simulation.

Public vs. graded

These notes, the examples, and the practice here are public and ungraded — study material only. No graded prompts, answer keys, rubrics, point values, or due dates appear on this site. Graded checkpoints, quizzes, homework, labs, the midterm, the project, and the final live in Blackboard (the LMS), which is authoritative for due dates, submissions, and grades. If this page and Blackboard ever disagree, follow Blackboard.

Looking ahead

Next week we condition. We have been computing probabilities of events over the whole sample space; in Week 3 — Conditional probability we ask what happens to those probabilities once we learn that part of the situation has already turned out a certain way — for example, that it is raining. The \(2 \times 2\) grid you built here is exactly the object we will slice: \(P(\text{on time} \mid \text{rain}) = 0.60\) lives inside the same table, and seeing why it differs from the marginal \(P(\text{on time}) = 0.81\) is the doorway to independence (Week 4) and Bayes’ rule (Week 5).