Disclaimer: Dieser Thread wurde aus dem alten Forum importiert. Daher werden eventuell nicht alle Formatierungen richtig angezeigt. Der ursprüngliche Thread beginnt im zweiten Post dieses Threads.

**Correction of Exercise 1.2.1 - Assignment 1**

Hello,

could someone explain why on the solution for this assignment P(A or B) = P(A) + P(B) - P(A)P(B)?

I mean, they are independent, right? Shouldn’t it be P(A or B) = P(A) + P(B)?

Also, can someone please fix on the pdf the solution for Exercise 1.2.2, it is going outside the document, so we cannot read the solution.

Thank you

Because that’s how the disjunction of two random variables behaves *in general*. It’s even one of Kolmogorov’s axioms.

Independence has nothing to do with that. If two variables are independent, then P(A and B)=P(A)*P(B) - maybe you’re thinking of that?

Apart from that, if P(A or B) = P(A) + P(B), then P(A and B)=0 - i.e. the two variables are *disjoint*, maybe you’re thinking of that

Will take care of that

Oh, I see. I had thought that disjoint and independent were the same concept. I understand now, thank you

Reading this got me a bit confused, so I looked it up: Kolmogorovs axiom is P(A or B) = P(A) + P(B) - P(A and B). It’s in the slides at 605 / theorem 2.19.

(Stochastic) Independence would then lead to P(A or B) = P(A) + P(B) - P(A)P(B).

It’s quite easy to imagine what happens here: say we have to overlapping sets A and B which obviously means A ∩ B ≠ {}. For counting all elements either contained in A or contained in B – formalized contained in A U B – of course we need |A| + |B|. But what we do here is counting those elements that are located in the cut set A ∩ B twice. And that is the reason why we need to subtract them once again. Hope I could clear your confusion a little bit