Axioms of Probability

Study guide: Exam P-1

The Axioms of Probability are three foundational rules that any valid Probability $P$ must satisfy:

Non-Negativity
Total Probability
Additivity

Axiom	Mathematical Statement	Description
1. Non-negativity	$P(E) \geq 0$	No negative odds. You cannot have a less-than-zero chance of an event occurring.
2. Total Probability	$P(S) = 1$	Something must happen. The probability of the entire sample space is 100%.
3. Additivity	$P\!\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} P(E_i)$	Add if no overlap. If events are Mutually Exclusive Events, the probability of "one or the other" is the sum of their individual probabilities.

Axioms satisfied? A probability model assigns $P(A) = 0.3$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.9$ where $A$ and $B$ are mutually exclusive. Does this violate the axioms of probability?

Answer If $A$ and $B$ are mutually exclusive, countable additivity requires $P(A \cup B) = P(A) + P(B) = 0.3 + 0.5 = 0.8$ . Since the model states $P(A \cup B) = 0.9 \neq 0.8$ , yes, it violates the third axiom.

Deriving the Complement Rule from the Axioms Using only the three axioms of probability, prove that $P(A^c) = 1 - P(A)$ .

Answer $A$ and $A^c$ are mutually exclusive (disjoint) and $A \cup A^c = S$ . By Axiom 3 (additivity): $P(A \cup A^c) = P(A) + P(A^c)$ But $A \cup A^c = S$ , so by Axiom 2: $P(S) = 1 = P(A) + P(A^c)$ Therefore $P(A^c) = 1 - P(A)$ . $\square$