Probability foundations - Exam P formula sheet

Every Exam P formula in probability foundations on one printable sheet, condensed from the 4 reference pages that teach them.

4 topics, 18 formula lines. Every line is condensed from the Exam P formula pages, which stay free to read in full. Print this page to get the sheet on paper; the site chrome drops away.

What is on this sheet

Probability foundations

Axioms and sets

Axioms
P(A)≥0,P(S)=1,P ⁣(⋃iAi)=∑iP(Ai) for disjoint AiP(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(Ac)=1−P(A)P(A^{c}) = 1 - P(A)
Inclusion-exclusion (two sets)
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
Inclusion-exclusion (three sets)
P(A∪B∪C)=∑P(A)−∑P(A∩B)+P(A∩B∩C)P(A\cup B\cup C) = \sum P(A) - \sum P(A\cap B) + P(A\cap B\cap C)
De Morgan
(A∪B)c=Ac∩Bc(A \cup B)^{c} = A^{c} \cap B^{c}

Counting

Multiplication rule
n1×n2×⋯×nkn_1 \times n_2 \times \dots \times n_k
Permutations
P(n,k)=n!(n−k)!P(n,k) = \dfrac{n!}{(n-k)!}order matters, no repetition
Combinations
(nk)=n!k! (n−k)!\binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}order does not matter
Multinomial
n!k1! k2!⋯km!\dfrac{n!}{k_1!\,k_2!\cdots k_m!}split n items into labelled groups
Stars and bars
(n+k−1k−1)\binom{n+k-1}{k-1}n identical items into k distinct boxes

Conditional probability

Definition
P(A∣B)=P(A∩B)P(B),P(B)>0P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}, \quad P(B) > 0
Multiplication rule
P(A∩B)=P(A∣B)P(B)P(A\cap B) = P(A\mid B)P(B)
Independence
P(A∩B)=P(A)P(B)  ⟺  P(A∣B)=P(A)P(A\cap B) = P(A)P(B) \iff P(A\mid B) = P(A)
Chain rule
P(A1∩⋯∩An)=P(A1)P(A2∣A1)⋯P(An∣A1∩⋯∩An−1)P(A_1\cap\dots\cap A_n) = P(A_1)P(A_2\mid A_1)\cdots P(A_n\mid A_1\cap\dots\cap A_{n-1})

Bayes' theorem

Law of total probability
P(A)=∑kP(A∣Bk)P(Bk)P(A) = \sum_{k} P(A\mid B_k)P(B_k)
Bayes
P(Bj∣A)=P(A∣Bj)P(Bj)∑kP(A∣Bk)P(Bk)P(B_j\mid A) = \dfrac{P(A\mid B_j)P(B_j)}{\sum_k P(A\mid B_k)P(B_k)}
Two-event form
P(B∣A)=P(A∣B)P(B)P(A∣B)P(B)+P(A∣Bc)P(Bc)P(B\mid A) = \dfrac{P(A\mid B)P(B)}{P(A\mid B)P(B) + P(A\mid B^{c})P(B^{c})}
Odds form
P(B1∣A)P(B2∣A)=P(A∣B1)P(A∣B2)⋅P(B1)P(B2)\dfrac{P(B_1\mid A)}{P(B_2\mid A)} = \dfrac{P(A\mid B_1)}{P(A\mid B_2)}\cdot\dfrac{P(B_1)}{P(B_2)}