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Thread: Prob. For Risk Mgmt 3-47 (Elements of Probability)

1. Originally Posted by jchaney
Thank you BeanCounter and NoMoreExams for your willingness to reply. I was not aware that P(b') is the complement to p(b). If you kindly refer to my initial post then you will see I did ask the question is there anything special about b' (B prime), or is just another variable. I've been under the premise that it treated another variable separate from A and B. I've reviewed the text I am studying have not crossed the point where B' is defined in such terms. Is P(b') the same as ~P(b). That notation is used in my text.

I can understand that why my most recent post seems so bizarre, but I would ask that for the respect of the forum you do not make insults to my intentions nor intellect. I believe this is the correct forum to ask such questions.

What part of the problem defines a and b as mutually exclusive or is it just assumed? The solution to this problem lists part of the step as P[aub]=p[a]+p[b]-p[anb], therefore I presumed that the events were not mutually exclusive.
I defined what P(B') means in my earlier post.

2. I'm surprised you didn't come across complements before you encountered this problem. You could also take the complements of the information given: If P[AuB] = 0.7, then the complement is P[AuB]' = 1 - P[AuB] = 1 - 0.7 = 0.3 = P[A'nB']. And the complement of P[AuB'] is P[A'nB] = 0.1.

If P[A] = P[AnB] + P[AnB'], then P[A'] = P[A'nB] + P[A'nB'].

P[A'] = 0.1 + 0.3 = 0.4

P[A] = 1 - P[A'] = 1 - 0.4 = 0.6.

3. Thank you RossMoney34 and NoMoreExams for your persistence and patience. I understand now.

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