chain rule

$$P(A_1,...,A_n) = P(A_1)P(A_2|A_1)P(A_3|A_1A_2)...P(A_n|A_1,...,A_{n-1})$$

$$P(A|B) = frac{P(B|A)P(A)}{P(B)}$$

Theorem: Let $A_1,A_2,…,A_n$ be the partition of the sample space $S$, with $P(A_1)>0$, Then :
$$P(B) = sum_i^n P(B|A_i)P(A_i)$$ Theorem: Let $A_1,A_2,…,A_n$ be the partition of the sample space $S$, for any event $B$ such that $P(B) > 0$, we have :
$$P(A_i|B) = frac{P(A_i)P(B|A_i)}{P(A_1)P(B|A_1)+dotsb+P(A_n)P(B|A_n)}$$

Conditional Probability is also the probability, so it inherent the property of probability (suppose the sample space is $S$):

• $P(S|E) = 1$ and $P(emptyset|E) = 0$
• if events $A_1,…$ are disjoint, then $P(cup_{j=1}^{infty}A_j|E) = sum_{j=1}^{infty}P(A_j|E)$
• $P(A^c|E) = 1 - P(A|E)$
• Inclusion-Exclusion : $P(Acup B|E) = P(A|E) + P(B|E) - P(Acap B|E)$

Theorem: Provided that $P(Acap E)>0$ and $P(Bcap E)>0$, we have:
$$P(A|B,E) = frac{P(B|A,E)P(A|E)}{P(B|E)}$$

Theorem: Let $A_1,A_2,…,A_n$ be the partition of the sample space $S$, with $P(A_i cap E) >0$, Then:
$$P(B|E) = sum_{i=1}^n P(B|A_i,E)P(A_i|E)$$

• $$P(A|B,C) = frac{P(A,B,C)}{P(B,C)}$$
• $$P(A|B,C) = frac{P(B|A,C)P(A|C)}{P(B|C)}$$
• $$P(A|B,C) = frac{P(C|A,B)P(A|B)}{P(A|C)}$$

Defination: Events $A$ and $B$ are independent if
$$P(Acap B) = P(A) P(B) Leftrightarrow P(A|B) = P(A) , P(B|A) = P(B)$$

• $A,B$ is disjoint : $P(Acap B) = 0$
• $A,B$ is independent : $P(A) = 0, P(B) = 0$

Defination: Events $A$ and $B$ are conditionally independent given E if:
$$P(Acap B|E) = P(A|E)P(B|E)$$ $$Contitional Independence nRightarrow Independence$$ $$Independence nRightarrow Contitional Independence$$