- #1

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Version 1

1. P(S)=1

2. P(A)>=0

3. If AnB=I, than P(AuB)=P(A)+P(B)

Version 2

1. P(S)=1

2. P(A)>=0

3. P(I)=0

4. P(AuB)=P(A)+P(B)-P(AnB)

It is easy to go from version 2 to version 1, and I can see how to show that P(I)=0 using version 1. The problem I'm having is how to derive axiom 4 of version 2 from version 1. I know I will have to consider two events made up of unions and intersections of A and B, to which I can apply axiom 3 of version 1, but I can't quite figure out how to do it. Thanks for any help.