What is Probability?

We all encounter uncertainty every day. Will it rain this afternoon? What are the chances a flipped coin lands heads? If you draw a card from a shuffled deck, how likely is it to be an ace? These questions all share a common structure: they ask about the likelihood of some outcome occurring in a situation we cannot predict with certainty.

Probability is the mathematical framework we use to quantify that uncertainty. It assigns a number between 0 and 1 to every possible outcome (or collection of outcomes) of a random process, where 0 means the outcome is impossible and 1 means it is certain. Everything in between tells us how likely an outcome is relative to all others.

Before we can assign probabilities, however, we need precise language for describing what we are measuring. The first part of this chapter is devoted entirely to building that language. It may seem like a lot of terminology upfront, but every definition here will be used constantly throughout the rest of these notes, so it is worth getting comfortable with these ideas early.

Prerequisites
These notes assume comfort with basic algebra and set notation (union, intersection, complement). If you have seen sets before, even briefly, you should be fine. No calculus is required for this chapter; it becomes useful in Chapter 2 when we deal with continuous random variables.

Experiments and Outcomes

The starting point for any probability problem is what we call a random experiment — a process whose outcome we cannot know in advance, but whose set of all possible outcomes we can describe completely.

Definition 1 — Random Experiment

A random experiment is any process that satisfies two conditions:

  1. All possible outcomes can be described (listed or characterized) before the experiment is performed.
  2. The actual outcome cannot be determined with certainty before the experiment is performed.

Common examples of random experiments include: flipping a coin, rolling a die, drawing a card from a shuffled deck, measuring the time until a light bulb fails, or recording the number of customers who arrive at a store in an hour. In each case we know what could happen, even if we do not know what will happen.

Each distinct result of an experiment is called an outcome or sample point. The collection of all possible outcomes is called the sample space.

Definition 2 — Sample Space

The sample space of a random experiment, denoted S (or sometimes Ω), is the set of all possible outcomes of that experiment. Each element of S is called a sample point or outcome.

Describing the sample space correctly is one of the most important steps in solving any probability problem, and it is a step that students frequently rush. A poorly defined sample space leads to errors that are very difficult to track down later, so take care here.

Example 1 — Flipping a Single Coin

A fair coin is flipped once. Describe the sample space.

Solution

There are two possible outcomes: the coin lands heads (H) or tails (T). Neither can happen at the same time, and there are no other possibilities. So the sample space is:

S = { H, T }

This sample space has 2 elements.

Example 2 — Rolling a Standard Die

A standard six-sided die is rolled once. Describe the sample space.

Solution

The die can land showing any of the faces 1 through 6. So:

S = { 1, 2, 3, 4, 5, 6 }

This sample space has 6 elements.

Example 3 — Flipping a Coin Twice

A fair coin is flipped twice. Describe the sample space, being careful to record the outcomes in order (first flip, second flip).

Solution

Each flip has two outcomes, so flipping twice gives 2 × 2 = 4 combinations. Listing them systematically:

S = { HH, HT, TH, TT }

Notice that HT and TH are different outcomes here because we record the order. HT means "heads on the first flip, tails on the second"; TH is the opposite. If order did not matter, HT and TH would be the same, and we would have only 3 outcomes—but for most probability calculations, keeping track of order is essential.

Events

When we ask a probability question, we are rarely interested in one specific outcome. More often we ask about a set of outcomes—"what is the probability of rolling an even number?" or "what is the probability that at least one flip is heads?" Such a collection of outcomes is called an event.

Definition 3 — Event

An event is any subset of the sample space S. Events are typically denoted by capital letters such as A, B, C, etc.

We say event A occurs if the actual outcome of the experiment is an element of A.

Two special events deserve names of their own:

  • The certain event is the entire sample space S itself. It always occurs, because whatever happens is by definition an element of S.
  • The impossible event is the empty set . It never occurs, because no outcome is an element of the empty set.
Example 4 — Events on a Die Roll

For the experiment "roll a standard six-sided die" with sample space S = {1, 2, 3, 4, 5, 6}, describe the following events in set notation:

  1. A = "the outcome is even"
  2. B = "the outcome is greater than 4"
  3. C = "the outcome is a multiple of 7"
Solution

(a) The even numbers in S are 2, 4, and 6:

A = { 2, 4, 6 }

(b) The numbers greater than 4 in S are 5 and 6:

B = { 5, 6 }

(c) There are no multiples of 7 in S, so:

C = ∅

C is the impossible event. No matter what the die shows, it cannot be a multiple of 7, so this event can never occur.

Assigning Probability

We now have the vocabulary to say what we are measuring. The next step is to actually assign numbers—probabilities—to events. There are several ways to build up to this, but the most general and rigorous starting point is the set of rules (called axioms) that any assignment of probabilities must satisfy.

Definition 4 — Probability (Kolmogorov Axioms)

A probability is a function P that assigns a real number to each event A in a sample space S such that the following three axioms hold:

  1. Non-negativity:  P(A) ≥ 0 for every event A.
  2. Normalization:  P(S) = 1.
  3. Additivity:  If A and B are mutually exclusive events (meaning AB = ∅), then
    P(AB) = P(A) + P(B).

Any function P satisfying these three axioms is a valid probability measure.

These axioms may look abstract. Think of them as ground rules: they do not tell you how to find the probability of an event—that depends on the problem—but they tell you what any valid answer must look like. Any formula or method you use to compute probabilities must produce answers consistent with these three rules. If it does not, something has gone wrong.

Classical Probability

For experiments where every outcome in the sample space is equally likely, there is a beautifully simple formula for computing probabilities.

Classical (Equally Likely Outcomes) Formula

If a sample space S consists of n equally likely outcomes, and an event A contains k of those outcomes, then:

P(A) = number of outcomes in A total number of outcomes in S  =  k n

This is the formula most people learn first when studying probability, and it works perfectly—as long as all outcomes truly are equally likely. The phrase "equally likely" is doing a lot of work here. A fair coin has equally likely outcomes; a bent coin does not. A well-shuffled deck has equally likely draws; a stacked deck does not. Always ask yourself whether the equally-likely assumption is justified before using this formula.

Example 5 — Rolling a Die (Revisited)

A fair six-sided die is rolled. Find the probability of each of the following events:

  1. A = "rolling an even number"
  2. B = "rolling a number greater than 4"
  3. C = "rolling a 3 or a 5"
Solution

The sample space is S = {1, 2, 3, 4, 5, 6} with n = 6 equally likely outcomes (since the die is fair).

(a) The even numbers in S are A = {2, 4, 6}, so k = 3.

P(A) = 3 6  =  1 2  = 0.5

(b) Numbers greater than 4 are B = {5, 6}, so k = 2.

P(B) = 2 6  =  1 3  ≈ 0.333

(c) C = {3, 5}, so k = 2.

P(C) = 2 6  =  1 3  ≈ 0.333

Notice that P(B) = P(C) even though the two events involve different numbers. Probability only depends on how many outcomes are in the event, not which ones.

Example 6 — Drawing a Card

One card is drawn at random from a standard 52-card deck. Find the probability that the card drawn is:

  1. A heart
  2. A face card (Jack, Queen, or King)
  3. A red ace
Solution

There are 52 cards in the deck, each equally likely to be drawn, so n = 52.

(a) There are 13 hearts in a standard deck.

P(heart) = 13 52  =  1 4  = 0.25

(b) There are 3 face cards per suit × 4 suits = 12 face cards.

P(face card) = 12 52  =  3 13  ≈ 0.231

(c) Aces come in 4 suits. The red suits are hearts and diamonds, so there are exactly 2 red aces.

P(red ace) = 2 52  =  1 26  ≈ 0.038

Basic Properties of Probability

From the three Kolmogorov axioms we can derive a number of useful properties. These are not new rules—they all follow logically from the axioms—but they come up so frequently that it is worth listing them explicitly.

Properties — Derived from the Axioms

For any events A and B in a sample space S:

  1. Complement rule:  P(Ac) = 1 − P(A)
  2. Impossible event:  P(∅) = 0
  3. Bounded:  0 ≤ P(A) ≤ 1 for every event A
  4. Addition rule:  P(AB) = P(A) + P(B) − P(AB)

Property 4 is called the General Addition Rule. When A and B are mutually exclusive, P(AB) = 0, and Property 4 reduces to Axiom 3.

Note — Complement Rule is Extremely Useful
The complement rule, P(Ac) = 1 − P(A), is one of the most frequently applied tools in probability. Many problems that look hard when phrased as "find the probability that A happens" become easy when rephrased as "find one minus the probability that A doesn't happen." Always consider whether the complement is easier to compute.

What's Coming

This section introduced the essential vocabulary and the classical probability formula. Here is a brief preview of how the rest of Chapter 1 develops these ideas:

  • 1.2 Sample Spaces & Events — We look more carefully at how to construct sample spaces for multi-step experiments, and how set operations (union, intersection, complement) translate into probability statements.
  • 1.3 Basic Probability Rules — We prove and apply the complement rule, the addition rule, and the inclusion-exclusion principle, with plenty of worked examples.
  • 1.4 Counting Techniques — For experiments with large sample spaces, listing every outcome is impractical. We develop systematic counting methods (permutations, combinations) that let us apply the classical formula without enumeration.
  • 1.5 Conditional Probability — How does partial information change our probability estimates? Conditional probability formalizes the idea of updating beliefs when we learn that some event has already occurred.
  • 1.6 Bayes' Theorem — One of the most celebrated results in all of mathematics. Bayes' theorem lets us "reverse" conditional probabilities and is the foundation of an entire approach to statistics.
  • 1.7 Independence — What does it really mean for two events to have nothing to do with each other? The formal definition of independence is more subtle than it first appears.

Practice Problems

Work through these before checking solutions. Problems are tagged Basic or Intermediate.

Problem 1
Basic

A fair coin is flipped and a fair six-sided die is rolled simultaneously.

  1. Write out the sample space S in full.
  2. How many outcomes does S contain?
  3. Let A = "the coin shows heads and the die shows an even number." List the outcomes in A.
Solution

(a) Each outcome is a pair (coin result, die result). Listing all:

S = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6),
   (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) }

(b) 2 × 6 = 12 outcomes.

(c) We need heads AND an even die: even die values are 2, 4, 6.

A = { (H,2), (H,4), (H,6) }
Problem 2
Basic

One integer is chosen at random from {1, 2, 3, …, 30}. Find the probability that the chosen number is:

  1. A multiple of 5
  2. A perfect square
  3. A multiple of 5 and a perfect square
Solution

Sample space: S = {1, 2, …, 30}, so n = 30. All outcomes equally likely.

(a) Multiples of 5 in S: {5, 10, 15, 20, 25, 30} → 6 outcomes.

P(multiple of 5) = 630  =  15  = 0.2

(b) Perfect squares in S: {1, 4, 9, 16, 25} → 5 outcomes.

P(perfect square) = 530  =  16  ≈ 0.167

(c) Must be both: multiples of 5 that are perfect squares → {25} only. 1 outcome.

P(mult. of 5 AND perfect square) = 130  ≈ 0.033
Problem 3
Basic

A bag contains 4 red balls, 6 blue balls, and 2 yellow balls. One ball is drawn at random.

  1. Find P(blue).
  2. Find P(not blue) using the complement rule.
  3. Find P(red or yellow) and verify it equals P(not blue).
Solution

Total balls: 4 + 6 + 2 = 12, so n = 12.

(a) 6 blue balls.

P(blue) = 612  =  12

(b) By the complement rule:

P(not blue) = 1 − P(blue) = 1 − 12  =  12

(c) Red or yellow = 4 + 2 = 6 balls.

P(red or yellow) = 612  =  12

This equals P(not blue), as expected: "red or yellow" is exactly the complement of "blue."

Problem 4
Intermediate

A fair six-sided die is rolled. Let:

  • A = "the result is even"
  • B = "the result is a prime number"
  1. List the outcomes in A, B, AB, and AB.
  2. Are A and B mutually exclusive? Explain.
  3. Find P(AB) using the general addition rule.
Solution

(a)

  • A = {2, 4, 6}
  • B = {2, 3, 5}  (primes on a die)
  • AB = {2}  (even AND prime)
  • AB = {2, 3, 4, 5, 6}

(b) No. A and B share the outcome 2, so AB ≠ ∅. They are not mutually exclusive.

(c) Using the general addition rule:

P(AB) = P(A) + P(B) − P(AB) = 36 + 3616 = 56

Check: AB = {2, 3, 4, 5, 6} has 5 elements, so P = 5/6. ✓

Problem 5
Intermediate

Three fair coins are flipped simultaneously.

  1. Write out the full sample space.
  2. Find the probability of getting exactly two heads.
  3. Find the probability of getting at least one head. (Hint: use the complement.)
Solution

(a) Each coin gives H or T, so 2 × 2 × 2 = 8 outcomes:

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

(b) Outcomes with exactly two heads: {HHT, HTH, THH} → 3 outcomes.

P(exactly 2 heads) = 38  = 0.375

(c) "At least one head" is the complement of "no heads at all." The only outcome with no heads is TTT.

P(at least 1 head) = 1 − P(no heads) = 1 − 18 = 78  = 0.875

Without the complement trick, you would have to count all 7 outcomes containing at least one H. The complement collapses that to a single calculation.

Problem 6
Intermediate

One card is drawn at random from a standard 52-card deck. Let:

  • A = "the card is a face card" (Jack, Queen, or King)
  • B = "the card is a black card" (clubs or spades)
  1. Find P(A), P(B), and P(AB).
  2. Find P(AB) using the general addition rule.
  3. What is the probability the card is neither a face card nor black?
Solution

(a)

  • Face cards: 3 per suit × 4 suits = 12, so P(A) = 12/52 = 3/13.
  • Black cards: clubs (13) + spades (13) = 26, so P(B) = 26/52 = 1/2.
  • Black face cards: J, Q, K of clubs + J, Q, K of spades = 6, so P(AB) = 6/52 = 3/26.

(b)

P(AB) = 1252 + 2652652 = 3252 = 813  ≈ 0.615

(c) "Neither face card nor black" is the complement of event AB:

P((AB)c) = 1 − 813 = 513  ≈ 0.385

These 20 remaining cards (52 − 32 = 20) are the non-face red cards: numbered red hearts and diamonds.

Problem 7
Intermediate

Suppose that P(A) = 0.45, P(B) = 0.30, and P(AB) = 0.15.

  1. Find P(AB).
  2. Find P(Ac) and P(Bc).
  3. Are A and B mutually exclusive? Justify your answer.
  4. Is it possible that P(AB) = 0.50 given the probabilities above? Why or why not?
Solution

(a)

P(AB) = 0.45 + 0.30 − 0.15 = 0.60

(b)

P(Ac) = 1 − 0.45 = 0.55    P(Bc) = 1 − 0.30 = 0.70

(c) No. Mutually exclusive requires P(AB) = 0. Here P(AB) = 0.15 > 0, so A and B can occur together.

(d) No. P(AB) cannot exceed either P(A) or P(B), since AB is a subset of both A and B. Here P(B) = 0.30, so the intersection probability cannot be 0.50.