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

$$P(A) \ge 0, \quad P(S) = 1, \quad P\!\left(\bigcup_i A_i\right) = \sum_i P(A_i) \text{ for disjoint } A_i$$

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.