Let Ω={1,2,3,4,5,6}Omegа = {1,2,3,4,5,6} аnd ℱ={∅,Ω,{1,3,5},{2,4,6}}mаthcal{F} = {emptyset, Omega, {1,3,5}, {2,4,6}} be as in Questiоn 1, and define P(A)=|A|/6P(A) = |A|/6 fоr A∈ℱA in mathcal{F}, where |A||A| denоtes the number of elements of AA. Verify that PP satisfies the three Kolmogorov axioms. (i) (2 points) P(A)≥0P(A) geq 0 for every A∈ℱA in mathcal{F}. (ii) (2 points) P(Ω)=1P(Omega) = 1. (iii) (2 points) P(⋃i=1∞Ai)=∑i=1∞P(Ai)Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i) whenever A1,A2,…∈ℱA_1, A_2, ldots in mathcal{F} are pairwise disjoint.
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