Solution-Kind of stuck and any help would be much

Kind of stuck and any help would be much appreciated

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If ACAC and BCBC are the events complementary to AA and BB, and p(A|B)>p(A)p(A|B)>p(A) and p(BC)>0p(BC)>0, prove that p(AC|BC)>p(AC)p(AC|BC)>p(AC).

Been trying to get my head around this but kind of stuck and any help would be much appreciated.

I know that p(A|B)=p(A and B)/p(B)p(A|B)=p(A and B)/p(B), p(AC)=1-p(A)p(AC)=1-p(A)... therefore p(A)=1-p(AC)p(A)=1-p(AC)... I

Reference no: EM131044810

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Kind of stuck and any help would be much appreciated : If ACAC and BCBC are the events complementary to AA and BB, and p(A|B)>p(A)p(A|B)>p(A) and p(BC)>0p(BC)>0, prove that p(AC|BC)>p(AC)p(AC|BC)>p(AC). Been trying to get my head around this but kind of stuck and any help would be much appreciated.

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Kind of stuck and any help would be much appreciated

Reference no: EM131044810

Reference no: EM131044810

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