AcademyProbability
Academy
Probability Axioms
Level 1 - Math II (Physics) topic page in Probability.
Principle
The probability axioms give the minimum rules that any finite probability model must satisfy. From them, common formulas such as complements and inclusion-exclusion follow rather than being separate assumptions.
Notation
\(S\)
sample space
\(A,B\)
events, each a subset of S
\(A^c\)
complement of A inside S
\(\varnothing\)
empty event
\(A\cup B\)
event that A or B occurs
\(A\cap B\)
event that both A and B occur
\(P(A)\)
probability of event A
Method
Derive the complement rule
Split the sample space
\[S=A\cup A^c\]
The pieces are disjoint
\[A\cap A^c=\varnothing\]
Use disjoint additivity
\[P(S)=P(A)+P(A^c)\]
Use certainty
\[1=P(A)+P(A^c)\]
Subtract P(A)
\[P(A^c)=1-P(A)\]
Derive the empty event rule
Add the empty event
\[S=S\cup\varnothing\]
The pieces are disjoint
\[S\cap\varnothing=\varnothing\]
Use disjoint additivity
\[P(S)=P(S)+P(\varnothing)\]
Subtract P(S)
\[P(\varnothing)=0\]
Derive inclusion-exclusion
Split the union into disjoint parts
\[A\cup B=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)\]
Add disjoint parts
\[P(A\cup B)=P(A\setminus B)+P(A\cap B)+P(B\setminus A)\]
Write P(A)
\[P(A)=P(A\setminus B)+P(A\cap B)\]
Write P(B)
\[P(B)=P(B\setminus A)+P(A\cap B)\]
Add P(A) and P(B)
\[P(A)+P(B)=P(A\setminus B)+P(B\setminus A)+2P(A\cap B)\]
Remove one overlap
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Rules
Non-negativity and upper bound
\[0\le P(A)\le 1\]
Certainty
\[P(S)=1\]
Disjoint additivity
\[A\cap B=\varnothing\quad\Rightarrow\quad P(A\cup B)=P(A)+P(B)\]
Complement
\[P(A^c)=1-P(A)\]
Empty event
\[P(\varnothing)=0\]
Inclusion-exclusion
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Examples
Question
If
\[P(A)=0.73\]
find \[P(A^c)\]
Answer
Use the complement rule:
\[P(A^c)=1-P(A)=1-0.73=0.27.\]
Checks
- Add probabilities directly only for disjoint events.
- For overlapping events, subtract the intersection once.
- The empty event has probability \(0\), but an event with probability \(0\) need not be impossible in every advanced model.
- If \(A\subseteq B\), then \(P(A)\le P(B)\).