Page Description Equation
65 conditional probability. provided $\Pr{ \left( B \right)} \neq0$ \[ \Pr{\left(A\middle| B\right)} =\frac{ \Pr{ \left(AB\right) }}{ \Pr{ \left(B\right) }}\\ \]
conditional probability \[ E_2 \subset E_1\\ \Rightarrow \Pr{ \left( E_2 \middle| A \right) } \le \Pr{ \left( E_1 \middle| A \right) } \]
conditional probability \[ \Pr{ \left( E \middle| A\right) } =1 - \Pr{\left( E^c \middle| A\right) } \]
96 conditional probability \[ \begin{multline*} \Pr{\left( E_1 \cup E_2 \middle| A \right)}\\ = \Pr{\left( E_1 \middle| A \right)} + \Pr{\left( E_2 \middle| A \right)} - \Pr{\left( E_1E_2 \middle| A \right)} \end{multline*} \]
68 multiplication rule for probability \[ \begin{multline*} \Pr{\left( A_1 \cdots A_n \right)}\\ =\Pr{\left( A_1 \right)} \Pr{\left( A_2 \middle| A_1 \right)} \Pr{\left( A_3 \middle| A_1A_2 \right)}\cdots\\ \cdots \Pr{\left( A_n \middle| A_1 \cdots A_{n-1} \right)} \end{multline*} \]
special cases of multiplication rule $\textbf{Case n=2}$ \[ \begin{align*} \Pr{\left( AB \right)} &= \Pr{\left( B \right)} \Pr{\left( A \middle| B \right)}\\ &= \Pr{\left( A \right)} \Pr{\left( B \middle| A \right)}\\ \end{align*} \] $\textbf{Case n=3}$ \[ \begin{multline*} \Pr{\left( A_1A_2A_3 \right)}\\ = \Pr{\left( A_1 \right)} \Pr{\left( A_2 \middle| A_1 \right)} Pr{\left( A_3 \middle| A_1A_2 \right)} \end{multline*} \]
Law of Total Probability. Holds for any event $A$ if $S = \bigcup_{i=1}^{n} B_i,$ $\Pr{\left( B_i \right)} > 0$ and $B_i B_j = \varnothing$ for all $i \neq j.$ \begin{align*} \Pr{\left( A \right)} &= \sum_{i=1}^{n} {\Pr{\left( B_i \right)} \Pr{\left( A | B_i \right)}} \\ &= \Pr{\left( B_1 \right)} \Pr{\left( A | B_1 \right)}\\ &\qquad+ \cdots + \Pr{\left( B_n \right)} \Pr{\left(A | B_n \right)} \end{align*}
recall that $A=AB\cup AB^C.$ \[ \begin{align*} \Pr{\left(A\right)} &= \Pr{\left( AB \cup A B^c \right)}\\ &= \Pr{\left( B \right)} \Pr{\left( A \middle| B \right)} + \Pr{\left( B^c \right)} \Pr{\left( A \middle| B^c \right)} \end{align*} \]
Holds for any collection of sets $S=\bigcup_{i=1}^{n}B_i, \Pr{\left(B_i\right)}>0,$ and $B_iB_j=\varnothing$ for all $i\neq j.$ \[ \Pr{\left( A \right)} = \sum_{i=1}^{n}{\Pr{\left( B_i \right)} \Pr{\left( A \middle| B_i \right)}} \]
77 Bayes' Formula. Holds for any collection of sets $S=\bigcup_{i=1}^{n}B_i,$ $\Pr{\left(B_i\right)}>0,$ and $B_iB_j=\varnothing$ for all $i\neq j.$ \[ \Pr{\left( B_j \middle| A \right)} = \frac{ \Pr{\left( B_j \right)} \Pr{\left( A \middle| B_j \right)} } { \sum_{i=1}^{n}{\Pr{\left(B_i\right)} \Pr{\left( A \middle| B_i \right)}} } \]
Bayes’ Formula Special Case. \[ \Pr{\left( A \right)} = \Pr{\left( A \middle| B\right)} \Pr{\left( B \right)} + \Pr{\left( A \middle| B^c \right)} \Pr{\left( B^c \right)} \]
86 $A_1,\ldots,A_n$ are independent events if for each $k,$ $1\le k\le n$ the statement holds. Otherwise, they are said to be dependent events. Mutually exclusive events are dependent. \[ \Pr{\left( A_{i_1} \cdots A_{i_k} \right)} =\Pr{\left( A_{i_1} \right)} \cdots \Pr{\left( A_{i_k} \right)} \]
84 If $A$ and $B$ are independent then so are $A$ and $B^c.$ \[ \Pr{\left( AB^c \right)} =\Pr{\left( A \right)} \Pr{\left( B^c \right)} \]
83 independent events when $n=2.$ Events $A$ and $B$ are independent if any of the three equations are true. And if one is true, all are, they are equivalent. \[ \begin{align*} \Pr\left(AB\right) &=\Pr\left(A\right)\Pr\left(B\right)\\ \Pr\left( A\middle| B \right) &= \Pr\left( A \right)\\ \Pr\left( B\middle| A \right) &= \Pr\left( B \right) \end{align*} \]
85 independent events when $n=3.$ Events $A,$ $B,$ $C$ are independent if all four statements are true at once. These are not equivalent statements so may not all be true at once. \[ \begin{align*} \Pr{\left( A \middle| B \right)} &= \Pr{\left( A \right)}\\ \Pr{\left( A \middle| C \right)} &= \Pr{\left( A \right)}\\ \Pr{\left(B\middle| C\right)} &=\Pr{\left(B\right)} \end{align*} \]
85 independent events when $n=3$ (alternate form). $A,$ $B,$ $C$ are independent events if the four equations are simultaneously true. \[ \begin{align*} \Pr{\left( ABC \right)} &=\Pr{\left( A \right)} \Pr{\left( B \right)} \Pr{\left( C \right)}\\ \Pr{\left( AB \right)} &= \Pr{\left( A \right)} \Pr{\left( B \right)}\\ \Pr{\left( AC \right)} &= \Pr{\left( A \right)} \Pr{\left( C \right)}\\ \Pr{\left( BC \right)} &= \Pr{\left( B \right)} \Pr{\left( C \right)}\\ \end{align*} \]
75 odds ratio \[ \frac{\Pr{\left( A \right)}} {\Pr{\left( A^c \right)}} = \frac{\Pr{\left( A \right)}}{1 - \Pr{\left( A \right)}} \]
95 Boole's inequality
Page Description

complete graph

deterministic

probabilistic method

Laplace's rule of succession

Theorem. If $\Pr{\left(A\right)}\gt0$ and $\Pr{\left(B\right)}\gt0$ and $\Pr{\left(A\right)}\lt\Pr{\left(A\middle| B\right)}$ then $\Pr{\left(B\right)}\lt\Pr{\left(B\middle| A\right)}$