A probability measure enjoys a number of basic properties.
In particular, the inversion principle states that the probability of the complement of an event is one minus the probability of the event. Another property allows us to compute the probability of a union of two events as the sum of the probabilities of the events minus the probability of their intersection.

In picture...

A simple random experiment

Consider the simple experiment in which we pick one out of three people at random, already used to illustrate the probability space.

The sample space is the set of three people: $\Omega = \{Teddy, Bob, John\}$.

The set of all possible events is
$\Sigma = \{$

$\emptyset$ (no one),

$\{Teddy\}$,

$\{Bob\}$,

$\{John\}$ ,

$\{Teddy, Bob\}$,

$\{Teddy, John\}$,

$\{Bob, John\}$,

$\{Teddy, Bob, John\}$ (everyone),

$\}$

For each of these events, the probability measure $P$ assigns a probability between 0 and 1. Assume that the probabilities of picking each individual are fixed: , and . Then, we can compute all the other probabilities from these ones via the inversion principle.

By definition, the probability that the outcome belongs to the sample space is
$$
P(\{Teddy, Bob, John\}) = P(\Omega ) = 1
$$

Hover over the list of events above to compute their probabilities

Union of non-disjoint events

Now click on events in the list to select them as part of a union of events

In maths...

Let $(\Omega, \Sigma, P)$ be a probability space. Then, we have the following properties.

Inversion principle

For any event $A \in \Sigma$, the probability of its complement $\overline{A} = \Omega \setminus A$ is
$$
P(\overline{A}) = 1- P(A).
$$

Proof: To show this, we use the definition of the probability measure regarding the probability of a union of disjoint events and the probability of the sample space as
$$
1 = P(\Omega) = P(A \cup \overline{A}) = P(A) + P(\overline{A}) .
$$

Union of events

For any events $A\in\Sigma$ and $B\in\Sigma$ (possibly non-disjoint), we have
$$
P(A\cup B ) = P(A) + P(B) - P(A\cap B).
$$

Proof: We combine
$$
P(A\cup B ) = P\left( (A \setminus (A\cap B) ) \cup B \right) = P( A \setminus (A\cap B) ) + P(B)
$$
with
$$
P(A) = P( A \setminus (A\cap B) ) + P(A\cap B) \quad \Rightarrow\quad P( A \setminus (A\cap B) ) = P(A) - P(A\cap B) .
$$