(i) Recall that two events "A_i" and "A_j" are independent iff. "P\\left(A_i\\cap A_j\\right)=P\\left(A_i\\right)P(A_j)". We may check this:
"P\\left(A_1\\cap {A_2}^c\\right)=P\\left(A_1\\cap \\left(A_2\\sqcup {A_2}^c\\right)\\right)-P\\left(A_1\\cap A_2\\right)\n=P\\left(A_1\\right)-P\\left(A_1\\right)P\\left(A_2\\right)=P\\left(A_1\\right)\\left(1-P\\left(A_2\\right)\\right)\n=P\\left(A_1\\right)\\left(P\\left(A_2\\sqcup {A_2}^c\\right)-P\\left(A_2\\right)\\right)\n=P\\left(A_1\\right)P\\left({A_2}^c\\right)."
(ii) We are given that set "{\\left\\{A_k\\right\\}}_{1,2,3}" is not just pairwise independent, it is mutually independent. Hence "A_1" and "A_2\\cap A_3" are independent. From (i), it is equivalent that "A_1" and "{A_2}^c\\cup {A_3}^c" are independent too.
Answer:
(i) Yes (if "{\\left\\{A_k\\right\\}}_{1,2,3}" is mutually or pairwise independent).
(ii) Yes (since "{\\left\\{A_k\\right\\}}_{1,2,3}" is mutually independent).
References:
1 "\\url{https:\/\/en.wikipedia.org\/wiki\/Independence_(probability_theory)#For_events}"
2 "\\url{https:\/\/en.wikipedia.org\/wiki\/Independence_(probability_theory)#More_than_two_events}"
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