some probability laws

$$begin{aligned} P [ S ] & = 1 \ P [ varnothing ] & = 0 \ P [ S backslash A ] & = 1 - P [ A ] \ P left[ A _ { 1 } cup A _ { 2 } right] & = P left[ A _ { 1 } right] + P left[ A _ { 2 } right] - P left[ A _ { 1 } cap A _ { 2 } right] end{aligned}$$

$$P left[ A _ { 1 } | A _ { 2 } right] = dfrac { left| A _ { 1 } cap A _ { 2 } right| } { left| A _ { 2 } right| } = dfrac { P left[ A _ { 1 } cap A _ { 2 } right] } { P left[ A _ { 2 } right] }$$

the conditional probability of the event A occurring, given that $B$ occurs as follows:

$$P [ B | A ] = dfrac { P [ A cap B ] } { P [ A ] }$$

two events A and B are independent if:

$P [ A cap B ] = P [ A ] P [ B ]$

$Updownarrow$

$begin{array} { l l } { P [ A | B ] = P [ A ] } & { text { if } P [ B ] neq 0 } \ { P [ B | A ] = P [ B ] } & { text { if } P [ A ] neq 0 } end{array}$

Partition. Let $S$ denote the sample space of some experiment, and consider $k$ events
$A _ { 1 } , ldots , A _ { k }$ in $S$ such that $ A _ { 1 } , ldots , A _ { k }$ are disjoint and $bigcup _ { i = 1 } ^ { k } A _ { i } = S .$ It is said that these
events form a partition of $S$.

Law of total probability. Suppose that the events $B _ { 1 } , ldots , B _ { k }$ form a partition of the
space $S$ and $operatorname { Pr } left( B _ { j } right) > 0$ for $j = 1 , ldots , k .$ Then, for every event $A$ in $S$,

$$operatorname { P } ( B ) = sumlimits _ { j = 1 } ^ { k } operatorname { P } left( A _ { j } right) operatorname { P } ( B | A _ { j } )$$

$$begin{align} P[A|B] &= dfrac{P(B|A) cdot P(A)}{P(B)}\ \ P[A_k | B ] &= dfrac{P[B cap A_k]}{P[B]} =dfrac{P[B | A_k] cdot P[A_k]}{ sumlimits_{j=1}^n P[B | A_j] cdot P[A_j]} end{align}$$