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 cs-401r:axioms-of-probability-theory [2014/09/05 16:55]cs401rpml [Third Axiom] cs-401r:axioms-of-probability-theory [2014/09/05 17:02] (current)cs401rpml [Third Axiom] Both sides previous revision Previous revision 2014/09/05 17:02 cs401rpml [Third Axiom] 2014/09/05 17:02 cs401rpml [Second Axiom] 2014/09/05 17:01 cs401rpml [First Axiom] 2014/09/05 17:01 cs401rpml [First Axiom] 2014/09/05 16:55 cs401rpml [Third Axiom] 2014/09/05 16:54 cs401rpml [Second Axiom] 2014/09/05 16:54 cs401rpml [First Axiom] 2014/09/05 14:10 ringger created Next revision Previous revision 2014/09/05 17:02 cs401rpml [Third Axiom] 2014/09/05 17:02 cs401rpml [Second Axiom] 2014/09/05 17:01 cs401rpml [First Axiom] 2014/09/05 17:01 cs401rpml [First Axiom] 2014/09/05 16:55 cs401rpml [Third Axiom] 2014/09/05 16:54 cs401rpml [Second Axiom] 2014/09/05 16:54 cs401rpml [First Axiom] 2014/09/05 14:10 ringger created Line 1: Line 1: == First Axiom == == First Axiom == The probability of any event $E$ is between 0 and 1: The probability of any event $E$ is between 0 and 1: - \begin{equation} + - 0 \leq P\left(E\right) \leq 1 + $0 \leq P\left(E\right) \leq 1$ - \end{equation} + == Second Axiom == == Second Axiom == The probability of the entire sample space $\Omega$ (equivalently,​ the "​certain event"​) is 1. The probability of the entire sample space $\Omega$ (equivalently,​ the "​certain event"​) is 1. - \begin{equation} + - P\left(\Omega\right) = 1 + $P\left(\Omega\right) = 1$ - \end{equation} + == Third Axiom == == Third Axiom == If for all $0 \leq i, j$ such that $i \neq j$ the events $A_i$ and $A_j$ are disjoint (i.e., $A_i \cap A_j = \emptyset$),​ then If for all $0 \leq i, j$ such that $i \neq j$ the events $A_i$ and $A_j$ are disjoint (i.e., $A_i \cap A_j = \emptyset$),​ then - \begin{equation} + - P\left(\bigcup_{j=1}^{\infty}{A_j}\right) = \sum_{j=1}^{\infty}{P\left(A_j\right)} + $P\left(\bigcup_{j=1}^{\infty}{A_j}\right) = \sum_{j=1}^{\infty}{P\left(A_j\right)}$ - \end{equation} + In the simple case of two variables, if $A \cap B = \emptyset$ then: In the simple case of two variables, if $A \cap B = \emptyset$ then: - \begin{equation} + - P\left(A \cup B\right) = P\left(A\right) + P\left(B\right) + $P\left(A \cup B\right) = P\left(A\right) + P\left(B\right)$ - \end{equation} + 