Conditional Probability: $P(A|B)$ - Probability of $A$ given $B$

• $P(A\cap B) = P(A|B)P(B)$
• $P(A|B)+P(A^c|B)=1$

Independent Events: $A,B$ are independent $\Leftrightarrow P(A|B)=P(A)$

Partition: $\{A_1, A_2, \dots, A_k\}$ is a partition of $\Omega$ if:

• $\Omega=A_1 \cup A_2 \cup \dots \cup A_k$
• $\forall i, j: i\ne j \Rightarrow A_i \cap A_j = \emptyset$
• (Intuitively, they’re all mutually exclusive events)

Law of Total Probability: If $\{B_1,B_2,\dots, B_k\}$ is a partition and $\forall i, P(B_i)>0$

• $\forall A, P(A)=\sum^k_{i=1}P(A|B_i)P(B_i)$
• $=P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+\dots+P(A|B_k)P(B_k)$

Bayes’ Theorem: If $\{B_1,B_2,\dots, B_k\}$ is a partition and $\forall i, P(B_i)>0$/m

• $P(B_j|A)=\frac{P(A|B_j)P(B_j)}{P(A)}$