Exam PProbability foundationsFree to read

Probability axioms and set identities

Three axioms generate every counting-free probability result on the exam, and inclusion-exclusion is the one identity that appears in some form on almost every sitting.

The formulas

Axioms

$$P(A)\ge 0,P(S)=1,P\left(\bigcup _i{A}_i\right)=\sum _{i}P(A_i)\text{ for disjoint }{A}_i$$

Complement

$$P(A^c)=1-P(A)$$

Inclusion-exclusion (two sets)

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

Inclusion-exclusion (three sets)

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

De Morgan

$$(A\cup B{)}^c=A^c\cap B^c$$

Where it comes from

1. Write $A\cup B$ as the disjoint union $A\cup (B\cap A^c)$; additivity gives $P(A\cup B)=P(A)+P(B\cap A^c)$.
2. Since $B=(B\cap A)\cup (B\cap A^c)$ is also disjoint, $P(B\cap A^c)=P(B)-P(A\cap B)$.
3. Substituting gives inclusion-exclusion — the overlap was counted twice, so it is subtracted once.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Add, add, subtract the overlap. Every 'at least one' question is that sentence.
• When a question says 'neither', take the complement of the union — De Morgan turns it into a single subtraction.

Traps

• Adding P(A) and P(B) without subtracting the intersection when the events are not disjoint.
• Assuming disjoint and independent are the same thing. Disjoint events with positive probability are never independent.

Related

• Conditional probability and independence
• Counting: permutations, combinations and partitions

Drill this: the Exam P question bank has original questions on this topic, and today’s free round is open to everyone.