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

[NoMoreExams]

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.

[RossMoney34]

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.

[jchaney]

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