Proof:Pr(S ∖ E) = 1 - Pr(E)

From CS2800 wiki
If Pr is a probability measure on S, then for all events [math]E [/math], [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(S \href{/cs2800/wiki/index.php/%E2%88%96}{∖} E) = 1 - \href{/cs2800/wiki/index.php/Pr}{Pr}(E) [/math]
Proof:
We know that [math]\href{/cs2800/wiki/index.php?title=Claim:A_%E2%88%A9_(B_%E2%88%96_A)_%3D_%E2%88%85&action=edit&redlink=1}{E ∩ (S ∖ E) = ∅} [/math]. By the third Kolmogorov axiom, this means that


[math]\begin{align*} \href{/cs2800/wiki/index.php/Pr}{Pr}(E) + \href{/cs2800/wiki/index.php/Pr}{Pr}(\href{/cs2800/wiki/index.php/S}{S} \href{/cs2800/wiki/index.php/%E2%88%96}{∖} E) &= \href{/cs2800/wiki/index.php/Pr}{Pr}(E \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} (\href{/cs2800/wiki/index.php/S}{S} \href{/cs2800/wiki/index.php/%E2%88%96}{∖} E)) && \href{/cs2800/wiki/index.php/Kolmogorov_axiom}{\text{by the third Kolmogorov axiom}} \\ &= \href{/cs2800/wiki/index.php/Pr}{Pr}(S) && \href{/cs2800/wiki/index.php?title=Claim:If_A_%E2%8A%86_B_then_B_%3D_A_%E2%88%AA_(B_%E2%88%96_A)&action=edit&redlink=1}{\text{since E ⊆ S}} \\ &= 1 && \href{/cs2800/wiki/index.php/Kolmogorov_axiom}{\text{by the second Kolmogorov axiom}} \end{align*} [/math]


Rearranging this equation gives the result.