Classical probability

A random experiment is one whose outcome cannot be predicted, although the set of all possibilities is known. That set is called the sample space and written $\Omega$; any subset of $\Omega$ we care about is an event.

For a roll of a die, $\Omega = \{1, 2, 3, 4, 5, 6\}$, and the event "an even number came up" is the subset $A = \{2, 4, 6\}$.

The classical formula

If all the elementary outcomes are equally likely, a probability is a ratio:

The numerator counts the outcomes favourable to $A$, the denominator counts all the possible outcomes. The equal-chances assumption matters: for a loaded die, or an urn holding balls of different sizes, this formula simply does not apply.

Properties

Three facts follow straight from the formula, and all three are worth treating as a check on your arithmetic:

A probability is never negative and never above one: the favourable set is part of the set of all possibilities. If you got something else, an outcome was counted twice or $\Omega$ was set up wrongly.

The extreme values have names of their own. An event of probability $0$ is impossible (a seven on one die), one of probability $1$ is certain (fewer than seven).

When Ω has to be counted

The real difficulty in these problems rarely lies in the formula and almost always in counting $|A|$ and $|\Omega|$ — that is, in the combinatorics of the previous lesson.

Take a roll of two dice. There are $6 \cdot 6 = 36$ outcomes (the rule of product), not eleven, even though the total takes only eleven values. That is exactly why the totals are not equally likely:

The total $7$ comes from six pairs: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. The total $12$ from just one: $(6,6)$.

The complementary event

The complement of $A$ is the event "$A$ did not happen", written $A'$. Since together they exhaust every possibility:

This is the most useful formula in the lesson, because the complement is very often easier to count than the event itself. The warning sign in a problem always reads the same: "at least one".

Probability as a percentage

It is often handier to quote a probability in percent — just multiply by $100\%$:

A probability of $\tfrac{3}{20}$ is $0.15$, that is $15\%$. The other way round: $40\%$ is $0.4$, that is $\tfrac{2}{5}$. The same number, a different notation — the fraction is handier to compute with, the percentage to talk about.

Combinatorics above and below the line

When an experiment draws several items at once, both counts have to come from the formulas of the previous lesson.

Frequency is not probability

A probability describes the model of an experiment, not what happened to come up. After thirty rolls of a die the outcomes are almost never evenly spread:

A bar of height zero says something quite different from a missing bar: the six could have come up, it just did not. Its probability stays $\tfrac{1}{6}$, and over a thousand rolls the frequencies would come noticeably closer to it — but over thirty they are under no obligation to.

Independent events

Two events are independent when one happening does not change the chances of the other. Then — and only then — the probabilities multiply:

Successive coin tosses are independent, so two heads have probability $\tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$. Drawing two balls without replacement is not: after the first draw the urn has changed, so the second probability has to be recomputed inside a smaller set of possibilities. A ratio computed inside a narrowed $\Omega$ like that is called a conditional probability and written $P(A|B)$.

Independence brings with it the commonest illusion in the topic: after five heads in a row, tails does not become more likely. A coin does not remember previous tosses.

Exercises

The three kinds of question match the three parts of the lesson. The $P(A)$ prompt gives the number of favourable outcomes $|A|$ and the number of all outcomes $|\Omega|$ — the answer is a fraction in lowest terms, so remember to reduce it. The $P(A')$ prompt gives the probability of an event and asks for its complement; that answer is a reduced fraction too. A prompt ending in an arrow and a percent sign asks you to convert the probability into percent — the answer is a number, for instance 15 for $\tfrac{3}{20}$.

Common mistakes

• Getting $\Omega$ wrong — with two dice there are 36 outcomes, not 11 and not 12.
• An unreduced fraction — a probability is quoted in lowest terms.
• Counting "at least one" directly — the complement is shorter and safer.
• Counting an outcome twice — favourable outcomes have to be listed so that none appears more than once.
• Multiplying probabilities of dependent events — $P(A)\cdot P(B)$ holds only under independence.
• The gambler fallacy — previous results do not change the chances of the next toss.
• The classical formula with unequal chances — a loaded die breaks the equal-likelihood assumption.