Union, intersection and complement of events

Events are sets

An event is a set of outcomes, so the language of sets carries over directly.

Words	Set notation	Meaning
$A$ or $B$ (or both)	$A \cup B$	at least one of them happens
$A$ and $B$	$A \cap B$	both happen
not $A$	$\overline{A}$	$A$ does not happen
$A$ and $B$ cannot both happen	$A \cap B = \varnothing$	mutually exclusive (disjoint)

In probability, "or" is inclusive unless the question says "exactly one" or "either … but not both".

The addition rule

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Adding $P(A)$ and $P(B)$ counts the overlap twice, once inside each, so it is subtracted once. Draw a Venn diagram with the overlap in the middle and the reason is visible. Only when $A$ and $B$ are mutually exclusive is the overlap empty, and then $P(A \cup B) = P(A) + P(B)$.

For three events the same idea gives

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$$

Complements and De Morgan

$P(\overline{A}) = 1 - P(A)$. The two rules that link complements with "and" and "or" are De Morgan's laws:

$$\overline{A \cup B} = \overline{A} \cap \overline{B}, \qquad \overline{A \cap B} = \overline{A} \cup \overline{B}.$$

In words: "not ($A$ or $B$)" is "neither"; "not ($A$ and $B$)" is "not both", which still allows one of them. So

$$P(\text{neither}) = 1 - P(A \cup B), \qquad P(\text{not both}) = 1 - P(A \cap B).$$

These two are easy to swap under time pressure, and a multiple-choice question may well offer both.

The four regions

Two events split the sample space into four regions: only $A$, only $B$, both, neither. Most questions give you three numbers and ask for a fourth, and the safest way to work is to fill in the four regions:

• both $= P(A \cap B)$
• only $A = P(A) - P(A \cap B)$, only $B = P(B) - P(A \cap B)$
• neither $= 1 - P(A \cup B)$
• exactly one $=$ only $A$ $+$ only $B$ $= P(A \cup B) - P(A \cap B)$

The four regions must add up to 1. Using a group of 100 people (or 1000) instead of decimals turns the regions into whole numbers, which is quicker without a calculator and makes mistakes visible.

Independent events

$A$ and $B$ are independent when knowing that one happened does not change the probability of the other. Then

$$P(A \cap B) = P(A) \times P(B).$$

Separate coin tosses, separate dice and draws with replacement are independent. Draws without replacement are not.

Independence is not the same as mutual exclusivity; for events with positive probability the two ideas are opposites. If $A$ and $B$ are mutually exclusive, knowing that $A$ happened tells you $B$ did not happen, which is as far from independence as it gets. Read the question for which one it states.

At least one, for independent events

The complement of "at least one happens" is "none happens", and for independent events "none" is a product:

$$P(\text{at least one}) = 1 - P(\overline{A})\,P(\overline{B})\cdots$$

For two events this agrees with the addition rule. If $P(A) = 0.3$ and $P(B) = 0.5$ independently, then $1 - 0.7 \times 0.5 = 0.65$, and $0.3 + 0.5 - 0.15 = 0.65$ too.

Divisibility questions

"A whole number from 1 to $N$ is chosen at random; what is the probability that it is divisible by 3 or 4?" is a union question. Count multiples of each, then subtract the multiples of both, which are the multiples of the least common multiple, here 12. A second check: the pattern repeats every 12 numbers, so count within one block of 12 and scale up.