The Mathematics of Maybe - Introduction to Probability

Will it rain before the school assembly ends? Will your team win tomorrow's hockey match? Will your name be pulled out of the box in the lucky draw? None of these questions has a certain answer in advance, yet none of them is a total mystery either — you already know the possible outcomes even though you cannot say which one will actually happen. That gap between "some outcome will happen" and "I cannot say which" is exactly what mathematics calls randomness, and probability is the branch of mathematics built to measure it precisely. This chapter turns the everyday habit of guessing — "it's unlikely to rain," "we're probably going to win" — into a numerical scale running from 0 (impossible) to 1 (certain), so that "unlikely" and "probably" can be compared and calculated with, exactly like lengths or areas. You will learn to describe every random experiment by its sample space, the complete list of things that could happen, and to describe any question about it — heads, an even number, a red ball — as an event, a chosen part of that sample space. From there, two independent routes to a number open up: theoretical probability, which reasons about favourable outcomes among all equally likely outcomes without ever touching a coin or a die, and experimental probability, which counts what actually happened across real trials, exactly the idea of relative frequency from statistics. A third setting, statistical probability, shows how a small sample can be used to estimate what an entire population is like. Along the way, tree diagrams give you a picture for experiments with more than one step, and the Law of Large Numbers explains why theoretical and experimental probability draw closer together the longer you keep tossing, rolling or sampling — while its most famous impostor, the Gambler's Fallacy, explains why a run of six heads in a row still leaves the next toss at exactly fifty-fifty.

Key takeaways

Probability measures the likelihood of a random event on a scale from 0 (impossible) to 1 (certain), written $P(E)$ with $0 \le P(E) \le 1$ .
A random experiment (tossing a coin, rolling a die) has an unpredictable individual outcome even though every possible outcome is known in advance.
The sample space $S$ is the complete list of possible outcomes of a random experiment, with sample size $n(S)$ ; an event $E$ is any part (subset) of $S$ .
Theoretical probability assumes all outcomes are equally likely: $P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$ .
Experimental probability is measured from real trials: $P(E) = \dfrac{\text{number of times } E \text{ occurred}}{\text{total number of trials}}$ , also called relative frequency.
Statistical probability extends the same idea from a sample to an entire population, letting a small survey estimate the whole group's behaviour.
The Law of Large Numbers says experimental probability drifts closer to theoretical probability as the number of trials grows.
The Gambler's Fallacy is the false belief that past independent outcomes change future ones — a fair coin and a fair die have no memory.
Tree diagrams list every outcome of a multi-step experiment as a path from a starting point, and multiplying probabilities along a path gives the probability of that combined outcome.
Key idea: probability turns "how likely" into a number between 0 and 1, computed either by reasoning about equally likely outcomes in a sample space (theoretical probability) or by counting what actually happened across trials (experimental probability), and the two agree more and more closely as the number of trials grows.

The core insight

Probability measures the likelihood of a random event on a scale from 0 to 1, either by reasoning about equally likely outcomes in a sample space (theoretical probability) or by counting relative frequency across repeated trials (experimental probability), and the Law of Large Numbers guarantees the two draw closer together as the number of trials grows.

What this chapter covers

The chapter opens by asking what makes an event random at all — a coin toss, a die roll, a lucky-draw slip — and introduces the probability scale, a number line from 0 (impossible) to 1 (certain) that turns vague words like "unlikely" into a precise measurement. It then presents two independent ways of pinning down that number objectively: experimental probability, built from counting actual trials (relative frequency), and theoretical probability, built from reasoning about equally likely outcomes without running any trial. A third strand, statistical probability, shows how the same relative-frequency idea, applied to a survey sample, estimates what an entire population is like. Along the way the chapter formalises sample space $S$ , sample size $n(S)$ , and event $E$ as a part of $S$ , and closes with tree diagrams for multi-step experiments, where probabilities along a path are multiplied together. Two ideas recur throughout: the Law of Large Numbers, and the Gambler's Fallacy — the mistaken belief that independent random events remember what happened before them.

The core ideas

Randomness and the probability scale

A random experiment (also called a random observation or trial) is an action that can be repeated, whose result cannot be predicted with certainty in advance, even though every possible outcome is known. Tossing a coin and rolling a die are the standard examples: a coin will land heads or tails, and a die will show one of 1 to 6, but no one can say in advance which a particular toss or roll will give. This unpredictability of a single trial, combined with full knowledge of what could happen, is exactly what randomness means in mathematics. Because probability compares wildly different situations — rain tomorrow, a hockey match, a lucky draw — on the same footing, it is measured on a fixed scale from 0 to 1.

0 \le P(E) \le 1,

where: $P(E)$ is the probability of an event $E$ ; $P(E) = 0$ means $E$ is impossible, $P(E) = 1$ means $E$ is certain, and every value strictly between 0 and 1 describes an event that is possible but not guaranteed — closer to 0 for "less likely," near 0.5 for "even chance," and closer to 1 for "more likely." Multiplying a probability by 100 converts it to a percentage chance, so $P(E) = 0.75$ and "75% chance" say exactly the same thing.

Figure

A number line from zero to one labelled impossible less likely even chance more likely and certain with five example events placed at their approximate positions

Sample space and events

The sample space, denoted $S$ , is the list of every possible outcome of a random experiment, with two firm rules: $S$ must include every outcome that could happen, and no outcome may be listed more than once. The number of outcomes in $S$ is the sample size, written $n(S)$ . Tossing one fair coin gives $S = \{H, T\}$ , $n(S) = 2$ ; rolling one fair die gives $S = \{1, 2, 3, 4, 5, 6\}$ , $n(S) = 6$ ; tossing two coins together gives $S = \{HH, HT, TH, TT\}$ , $n(S) = 4$ , because $HT$ (first coin heads, second tails) and $TH$ (first tails, second heads) are genuinely different outcomes. An event is any one outcome or group of outcomes chosen out of the sample space — formally, an event is a subset of $S$ .

E \subseteq S, \qquad n(E) \le n(S).

where: $E$ is the event (for example, "at least one coin shows heads"), $n(E)$ is the number of outcomes belonging to $E$ , and $n(S)$ is the total sample size. For two coins, the event "at least one head" is $E = \{HH, HT, TH\}$ , so $n(E) = 3$ out of $n(S) = 4$ ; the event "the number rolled is greater than 4" on a single die is $E = \{5, 6\}$ , so $n(E) = 2$ out of $n(S) = 6$ .

Theoretical probability

Theoretical probability answers "what should happen in a perfectly fair situation?" without performing any experiment at all. It assumes every outcome in the sample space is equally likely — no reason exists to believe any one outcome is more likely than another — and simply compares the count of favourable outcomes to the total count of possible outcomes.

P(E) = \frac{n(E)}{n(S)} = \frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}. \qquad (1)

where: $n(E)$ counts the outcomes that make the event happen, and $n(S)$ counts every possible outcome. Rolling a fair die, the probability of getting a 4 is $P(4) = \dfrac{1}{6} \approx 0.167$ , since exactly one face shows a 4 out of six equally likely faces. Picking a random letter from the word PROBABILITY (11 letters, with the letter B appearing twice), $P(B) = \dfrac{2}{11} \approx 0.182$ — here the "outcomes" are the 11 letter-positions, and repeated letters simply mean more than one favourable position.

Experimental probability and the Law of Large Numbers

Experimental probability answers a different question: "what actually happened, based on real trials?" It requires performing the experiment, or gathering data from past observations, and is calculated as a relative frequency — the fraction of trials in which the event occurred.

P(E) = \frac{\text{number of times } E \text{ occurred}}{\text{total number of trials}}. \qquad (2)

where: the numerator is a count taken directly from recorded trials, and the denominator is the total number of trials performed. Rolling a die 50 times and getting a 4 exactly 8 times gives an experimental probability of $\dfrac{8}{50} = 0.16$ — close to, but not equal to, the theoretical value $\dfrac{1}{6} \approx 0.167$ for the same event, which is normal for a small number of trials. The Law of Large Numbers states that as trials increase, experimental probability tends to get closer to theoretical probability; rolling the die 6000 times instead of 50 would typically bring the relative frequency of 4s much nearer to $\dfrac{1}{6}$ . Experimental probability never contradicts theoretical probability — it converges toward it.

Statistical probability: from a sample to a population

A third, closely related setting uses relative frequency from a survey of many different individuals rather than repeats of one experiment — this is statistical probability, widely used in business, insurance and research. Suppose a survey of 50 students finds that 20 like mango best; the estimated probability that a randomly chosen student likes mango is $\dfrac{20}{50} = 0.4$ . Its real power appears in estimating a much larger population: if the whole school has 1500 students and the class of 50 is a fair (unbiased) sample of it, the same proportion estimates that about $0.4 \times 1500 = 600$ students in the school prefer mango. A larger, more representative sample — drawn from many classes rather than one — makes such estimates more trustworthy; choosing a fair, unbiased sample is itself a branch of statistics called sampling.

Tree diagrams for multi-step experiments

A multi-step experiment performs a series of trials one after another — tossing a coin twice, or rolling a die and then tossing a coin. A tree diagram is the standard picture for such an experiment: starting from a single point, branches are drawn to each outcome of the first step; from the end of each of those branches, further branches are drawn to each outcome of the second step, and so on. Every complete path from the start to the end of the tree is exactly one outcome of the combined sample space, and writing the probability on each branch lets you find the probability of any complete path by multiplying along it.

P(\text{a complete path}) = P(\text{branch}_1) \times P(\text{branch}_2) \times \cdots

where: each factor is the probability of the branch chosen at that step (for a fair coin, $\tfrac{1}{2}$ at every branch). Tossing two fair coins, the path root, first-toss-heads, second-toss-heads has $P(HH) = \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} = 0.25$ — matching the theoretical-probability count $\tfrac{1}{n(S)} = \tfrac{1}{4}$ found directly from the sample space $S = \{HH, HT, TH, TT\}$ .

Figure

A tree diagram for tossing a fair coin twice with two first branches for heads and tails each labelled probability one half and four leaf outcomes HH HT TH TT each labelled probability one quarter

Derivation AThe Fundamental Counting Principle for tree diagrams

The goal is to find a general rule for the sample size $n(S)$ of a multi-step experiment built from several independent steps — tossing two coins, rolling a die and then tossing a coin, drawing a ball twice with replacement — without drawing every branch of a tree diagram by hand once an experiment has more than two or three steps. The key idea is that a tree diagram is really a picture of repeated multiplication: at each new step, every path already drawn splits into as many new branches as there are outcomes at that step.

Step 1. Start with one step: tossing a single fair coin has 2 outcomes, so $S_1 = \{H, T\}$ , $n(S_1) = 2$ — a single starting point splitting into 2 branches.

Step 2. Attach a second coin toss to each of those 2 branches. Each first-toss branch splits again into 2 further branches, so the total number of complete paths is $2 \times 2 = 4$ , matching the sample space already listed for two coins, $S = \{HH, HT, TH, TT\}$ .

Step 3. Generalise. If the first step has $n_1$ possible outcomes and the second (performed after the first, regardless of its result) has $n_2$ possible outcomes, then every one of the $n_1$ first branches splits into $n_2$ further branches, giving $n_1 \times n_2$ complete paths. A third step with $n_3$ outcomes multiplies the count again, and so on for any number of steps: $n(S) = n_1 \times n_2 \times n_3 \times \cdots$ — the Fundamental Counting Principle.

Step 4. Combine this with theoretical probability. If every outcome at every step is equally likely, every complete path through the tree is equally likely too, so each carries probability $\dfrac{1}{n(S)}$ , exactly the product of the probability written on each branch along that path. For two fair coins, $P(HH) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ , matching $\dfrac{1}{n(S)} = \dfrac{1}{4}$ . Rolling a die and then tossing a coin gives $n(S) = 6 \times 2 = 12$ equally likely outcomes, each with probability $\dfrac{1}{12}$ , without writing out all twelve branches.

Go deeper · Derivation B · 1 minThe complement rule for probability

Derivation BThe complement rule for probability

The goal is to prove a shortcut that saves counting work in many problems: instead of directly counting the favourable outcomes for an event $A$ , it is often quicker to count the outcomes where $A$ does not happen, and subtract that fraction from 1. This matters whenever "not $A$ " is a much shorter list than " $A$ itself" — "not red" among several colours is usually easier to picture directly than adding up every non-red colour one at a time. The proof rests on nothing beyond the definition of theoretical probability and the fact that every outcome in a sample space belongs to exactly one of two groups: those in $A$ , and those not in $A$ .

Step 1. Let $S$ be the sample space with $n(S)$ equally likely outcomes, and let $A$ be an event with $n(A)$ favourable outcomes. Define "not $A$ ," written $A'$ , as every outcome of $S$ that is not in $A$ . Because every outcome of $S$ is either in $A$ or not in $A$ , and never both, the two events $A$ and $A'$ together account for the whole of $S$ , with no outcome shared and none left out.

Step 2. Count outcomes on both sides. Since $A$ and $A'$ split $S$ into two non-overlapping groups making up all of $S$ , their counts add to the total: $n(A) + n(A') = n(S)$ .

Step 3. Divide every term by $n(S)$ , and apply $P(E) = \dfrac{n(E)}{n(S)}$ to each piece: $\dfrac{n(A)}{n(S)} + \dfrac{n(A')}{n(S)} = \dfrac{n(S)}{n(S)}$ , which is precisely $P(A) + P(A') = 1$ .

Step 4. Rearrange to isolate $P(A')$ , giving the complement rule:

P(A') = 1 - P(A).

Apply it to a bag of 3 red, 2 blue and 1 green ball, 6 in total: $P(\text{red}) = \dfrac{3}{6} = \dfrac{1}{2}$ , so without recounting anything, $P(\text{not red}) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$ — matching a direct count of the 3 non-red balls out of 6.

Worked examples

Board level - the letter B and a vowel in PROBABILITY

A letter is picked at random from the word PROBABILITY. Find (a) the probability of picking the letter B, and (b) the probability of picking a vowel.

Step 1. List the letters: P, R, O, B, A, B, I, L, I, T, Y — 11 letters, so $n(S) = 11$ . B appears twice, so for part (a), $n(B) = 2$ .

Step 2. Apply theoretical probability for part (a): $P(B) = \dfrac{n(B)}{n(S)} = \dfrac{2}{11} \approx 0.182$ or $18.2\%$ .

Step 3. For part (b), the vowels among the 11 letters are O, A, I, I, so $n(\text{vowel}) = 4$ , and $P(\text{vowel}) = \dfrac{4}{11} \approx 0.364$ or $36.4\%$ . Answer: $P(B) = \dfrac{2}{11} \approx 0.182$ , and $P(\text{vowel}) = \dfrac{4}{11} \approx 0.364$ .

Board level - two coin tosses by tree diagram

Two fair coins are tossed together. Using a tree diagram, find (a) the sample space, (b) the probability of getting two heads, and (c) the probability of getting exactly one head.

Step 1. Draw the tree: the first coin branches into H and T, each probability $\tfrac{1}{2}$ ; from each, the second coin again branches into H and T, each probability $\tfrac{1}{2}$ . The four complete paths give $S = \{HH, HT, TH, TT\}$ , $n(S) = 4$ .

Step 2. Multiply along the HH path for part (b): $P(HH) = \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} = 0.25$ .

Step 3. For part (c), "exactly one head" is $E = \{HT, TH\}$ — the two paths with one head and one tail; $HH$ has two heads and $TT$ has none, so neither belongs to $E$ . So $n(E) = 2$ , and $P(E) = \dfrac{2}{4} = 0.5$ . Answer: $S = \{HH, HT, TH, TT\}$ ; $P(\text{two heads}) = 0.25$ ; $P(\text{exactly one head}) = 0.5$ .

Higher-order thinking (HOTS / olympiad-style) - the sum of two dice is a prime greater than 5

Two fair six-sided dice are rolled together. What is the probability that the sum of the two numbers shown is a prime number greater than 5?

Step 1. Find the sample size. Each die has 6 equally likely faces and the two rolls are independent steps, so by the Fundamental Counting Principle $n(S) = 6 \times 6 = 36$ equally likely ordered pairs (first die, second die) — $(1,6)$ and $(6,1)$ are different outcomes.

Step 2. List the possible sums, ranging from $2$ ( $1+1$ ) to $12$ ( $6+6$ ), and pick out the primes greater than 5: only $7$ and $11$ qualify — $2$ is prime but too small, and $5$ is excluded by "greater than 5."

Step 3. Count the ordered pairs for each qualifying sum. Sum $= 7$ : $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — 6 outcomes. Sum $= 11$ : $(5,6), (6,5)$ — 2 outcomes. Total favourable outcomes: $6 + 2 = 8$ .

Step 4. Apply theoretical probability: $P(\text{sum is a prime greater than 5}) = \dfrac{8}{36} = \dfrac{2}{9} \approx 0.222$ . Answer: $P = \dfrac{2}{9} \approx 0.222$ or $22.2\%$ . Timing note: about two to three minutes if the 6 by 6 grid of sums is drawn out first rather than listed pair by pair. The two traps are treating $(1,6)$ and $(6,1)$ as the same outcome, which wrongly shrinks $n(S)$ below 36, and forgetting to test every candidate sum against both conditions — prime, and strictly greater than 5 — since the sum 2 is prime but too small, so each sum from 2 to 12 needs checking against both conditions before being counted.

Common traps

Treating every experiment as having equally likely outcomes is a very common error. A driver starting a car either succeeds or fails — two outcomes — but these are not equally likely the way heads and tails are; the formula $P(E) = n(E)/n(S)$ only applies once the outcomes are known, or stated, to be equally likely. Miscounting the sample space is another frequent slip: forgetting an outcome, or listing one twice, silently changes $n(S)$ and throws off every probability built from it. Write the full sample space out explicitly, in curly brackets, before calculating anything. Confusing frequency with relative frequency causes errors in statistics-flavoured questions: "15 out of 50 students like football" is a frequency (a raw count), while the relative frequency, and the estimated probability, is $\dfrac{15}{50} = 0.3$ — check whether a question wants the count or the fraction. Expecting a small number of trials to exactly reproduce the theoretical probability is another misunderstanding: rolling a die 12 times and getting a 3 only twice, rather than matching $\tfrac{1}{6} \times 12 = 2$ perfectly, is entirely normal — small samples wobble, and only the Law of Large Numbers, over many more trials, closes the gap. In tree-diagram problems, students sometimes add probabilities along a single path instead of multiplying them, or multiply across two different paths instead of adding those paths together; the rule is multiply along one path, add across separate qualifying paths. Finally, checking that a probability answer lies between 0 and 1 is a useful last step: an answer like $\dfrac{7}{4}$ or $-0.2$ signals an arithmetic slip before you need to recheck every earlier step.

Common trapThe Gambler's Fallacy — believing a fair coin or die remembers its past outcomes

The single biggest misconception in this chapter is the belief that if something random has happened several times in a row, the opposite is somehow "due" to balance things out. Six heads in a row on a fair coin does not make tails more likely on the seventh toss; the coin has no memory, and each toss is an independent event with $P(\text{heads}) = P(\text{tails}) = 0.5$ , exactly as before. The same applies to a die: rolling three 6s in a row does not lower the probability of a fourth 6, which stays fixed at $\tfrac{1}{6} \approx 0.166$ on every single roll. This mistaken reasoning is called the Gambler's Fallacy. What the Law of Large Numbers actually guarantees is different and much weaker: over a very large number of tosses, the overall proportion of heads will settle close to 0.5 — but it does this by swamping the earlier run with a large number of future tosses, not by the coin "correcting" itself on any particular next toss. Probability describes long-run behaviour, never the single next outcome.

You should be able to…

explain why a random experiment is unpredictable on a single trial even though every possible outcome is known in advance
place any event on the probability scale from 0 (impossible) to 1 (certain), using words like less likely, even chance and more likely
write the sample space $S$ of a random experiment correctly, with no outcome missing or repeated, and state its sample size $n(S)$
identify an event $E$ as a part of a sample space and count $n(E)$
calculate theoretical probability $P(E) = n(E)/n(S)$ for equally likely outcomes, including outcomes built from letters of a word
calculate experimental probability (relative frequency) from the results of real trials, and distinguish frequency from relative frequency
use a sample survey to estimate a proportion for a whole population, and explain what makes a sample more trustworthy
draw a tree diagram for a two-step experiment, read off its sample space, and multiply along a path to find the probability of a combined outcome
state and apply the complement rule $P(\text{not } A) = 1 - P(A)$
explain the Law of Large Numbers and recognise the Gambler's Fallacy as a misunderstanding of independent events

Ask the tutor: Pick any everyday random event from your own day, describe its sample space, and ask the tutor to check your theoretical and experimental probability calculations step by step.

What is the actual difference between experimental probability and theoretical probability?

Theoretical probability is calculated purely by reasoning, without performing any trial: assume all outcomes are equally likely and divide favourable outcomes by all possible outcomes, as in $P(4) = 1/6$ for a fair die. Experimental probability instead comes from real data — toss the coin, roll the die, or gather survey results, and divide how many times the event happened by the total number of trials. The two give close but slightly different numbers for a small number of trials; the Law of Large Numbers is why experimental probability gets closer to theoretical probability as trials increase.

If a coin has landed heads six times in a row, isn't tails more likely on the next toss?

No — this is the Gambler's Fallacy, the most common mistake in this chapter. A fair coin has no memory of previous tosses; each toss is independent, so the probability of heads on the seventh toss is exactly the same $0.5$ it always was, regardless of the six heads before it. What does change over a very large number of tosses is the overall proportion of heads, which tends to settle near 0.5 — but that happens because future tosses are numerous enough to outweigh an early run, not because the coin corrects itself for what already happened.

Why does the sample space use curly brackets, and why can an outcome not be listed twice?

Curly-bracket notation, such as $S = \{H, T\}$ , is the standard way mathematics writes a set — a collection of distinct items with no particular order. A sample space must list every outcome exactly once because $n(S)$ is the denominator in $P(E) = n(E)/n(S)$ ; an outcome listed twice makes $n(S)$ too large, so every probability from that sample space comes out too small. Listing each outcome exactly once, with nothing left out, keeps $n(S)$ accurate.

How is estimating a population from a survey sample different from tossing a coin many times?

Both use the same relative-frequency idea, applied to different kinds of repetition. Tossing a coin 50 times repeats the same experiment 50 times, and the relative frequency estimates that experiment's probability. Surveying 50 students about their favourite fruit instead asks the same question of 50 different individuals, and the relative frequency (say, 20 out of 50 liking mango) estimates a proportion in the wider population — for example, all 1500 students in the school. The arithmetic is identical; what differs is that a survey's accuracy also depends on the sample being large and fairly representative, not biased toward one group.