Several theorems in probability can be deduced from the axioms of probability. These theorems can be applied to calculate probabilities that we may desire to know. One such result is known as the complement rule. This statement allows us to calculate the probability of an event *A* by knowing the probability of the complement *A*^{C}. After stating the complement rule, we will see how this result can be proved.

## The Complement Rule

The complement of the event *A* is denoted by *A*^{C}. The complement of *A* is the set of all elements in the universal set, or sample space S, that are not elements of the set *A*.

The complement rule is expressed by the following equation:

P(*A*^{C}) = 1 – P(*A*)

Here we see that the probability of an event and the probability of its complement must sum to 1.

## Proof of the Complement Rule

To prove the complement rule, we begin with the axioms of probability. These statements are assumed without proof. We will see that they can be systematically used to prove our statement concerning the probability of the complement of an event.

- The first axiom of probability is that the probability of any event is a nonnegative real number.
- The second axiom of probability is that the probability of the entire sample space
*S*is one. Symbolically we write P(*S*) = 1. - The third axiom of probability states that If
*A*and*B*are mutually exclusive ( meaning that they have an empty intersection), then we state the probability of the union of these events as P(*A*U*B*) = P(*A*) + P(*B*).

For the complement rule, we will not need to use the first axiom in the list above.

To prove our statement we consider the events *A*and *A*^{C}. From set theory, we know that these two sets have empty intersection. This is because an element cannot simultaneously be in both *A* and not in *A*. Since there is an empty intersection, these two sets are mutually exclusive.

The union of the two events *A* and *A*^{C} are also important. These constitute exhaustive events, meaning that the union of these events is all of the sample space *S*.

These facts, combined with the axioms give us the equation

1 = P(*S*) = P(*A* U *A*^{C}) = P(*A*) + P(*A*^{C}) .

The first equality is due to the second probability axiom. The second equality is because the events *A* and *A*^{C} are exhaustive. The third equality is because of the third probability axiom.

The above equation can be rearranged into the form that we stated above. All that we must do is subtract the probability of *A* from both sides of the equation. Thus

1 = P(*A*) + P(*A*^{C})

becomes the equation

P(*A*^{C}) = 1 – P(*A*).

Of course, we could also express the rule by stating that:

P(*A*) = 1 – P(*A*^{C}).

All three of these equations are equivalent ways of saying the same thing. We see from this proof how just two axioms and some set theory go a long way to help us prove new statements concerning probability.