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

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

    I have a couple of questions on this problem.
    P[AuB]=0.7 and P[AuB']=0.9
    Determine P[A]

    I understand the substitution via Disjunction Rule to:

    P[AuB]=P[A]+P[B]-P[AnB]
    P[AuB']=P[A]+P[B']-P[AnB']

    Then combining the terms to:

    P[AuB]+P[AuB']=2P[A]+(P[B]+P[B'])-(P[AnB]+P[AnB'])

    But, the next part seems like hand waving on the right-hand side of the equation.

    0.7+0.9=2P[A]+1-P[(AnB)u(AnB')]

    How do we get to 1-P[(AnB)u(AnB')] from (P[B]+P[B'])-(P[AnB]+P[AnB'])?

    The next step suggests:

    1.6=2P[A]+1-P[A] and subsequently P[A]=0.6

    Also, is there any notable difference between B' (B prime?) and say C or any other variable. Thanks in advance.

  • #2
    I avoided that messy equation altogether. Try taking the complements of AuB and AuB', it's a much smoother route.

    Comment


    • #3
      Originally posted by jchaney View Post
      i have a couple of questions on this problem.
      P[aub]=0.7 and p[aub']=0.9
      determine p[a]

      i understand the substitution via disjunction rule to:

      P[aub]=p[a]+p[b]-p[anb]
      p[aub']=p[a]+p[b']-p[anb']

      then combining the terms to:

      P[aub]+p[aub']=2p[a]+(p[b]+p[b'])-(p[anb]+p[anb'])

      but, the next part seems like hand waving on the right-hand side of the equation.

      0.7+0.9=2p[a]+1-p[(anb)u(anb')]

      how do we get to 1-p[(anb)u(anb')] from (p[b]+p[b'])-(p[anb]+p[anb'])?

      The next step suggests:

      1.6=2p[a]+1-p[a] and subsequently p[a]=0.6

      also, is there any notable difference between b' (b prime?) and say c or any other variable. Thanks in advance.
      (p[b]+p[b'])=1

      Comment


      • #4
        Really?

        Originally posted by BeanCounter View Post
        (p[b]+p[b'])=1
        Why must (p[b]+p[b'])=1? I don't think this is the case. If so, please kindly support.

        Comment


        • #5
          Originally posted by jchaney View Post
          Why must (p[b]+p[b'])=1? I don't think this is the case. If so, please kindly support.
          By definition P(B') = 1 - P(B) therefore P(B) + [1 - P(B)] = 1

          Think about it logically, probability of an event happening OR probability of an even NOT happening HAS TO BE 100% i.e. either it does or does not happen.

          Comment


          • #6
            Hmm?

            If one has only three choices for breakfast combination.
            A = Toast
            B = Eggs
            and B' = Ham

            Then we could say that there is a 70% chance of either toast or eggs for breakfast, and 90% chance of toast or ham. Further 30% not toast nor eggs or 10% not toast nor ham. But, what of the those who have toast and eggs, or toast and ham. By this we are saying those had ham are equal to everyone less those who had eggs.

            I get P(a) + p(b) - P(anb) = 1 and P(a) + p(b') - P(anb') = 1
            but, for p(b)+ p(b') = 1 would not p(anb) + p(a) equal to p(b') and consequently p(a) + p(anb') = p(b)? Where is my logic flawed?

            Comment


            • #7
              Originally posted by jchaney View Post
              If one has only three choices for breakfast combination.
              A = Toast
              B = Eggs
              and B' = Ham

              Then we could say that there is a 70% chance of either toast or eggs for breakfast, and 90% chance of toast or ham. Further 30% not toast nor eggs or 10% not toast nor ham. But, what of the those who have toast and eggs, or toast and ham. By this we are saying those had ham are equal to everyone less those who had eggs.

              I get P(a) + p(b) - P(anb) = 1 and P(a) + p(b') - P(anb') = 1
              but, for p(b)+ p(b') = 1 would not p(anb) + p(a) equal to p(b') and consequently p(a) + p(anb') = p(b)? Where is my logic flawed?
              In your example:
              Equation 1: P(a) + p(b) - P(anb) = 1
              Equation 2: P(b)+ P(b') = 1

              Then, we get P(b') = P(a) - P(anb) but not p(anb) + p(a)

              Test:
              If a and b are mutually exclusive, P(anb) = 0. There two equations are still consistent.

              Your another example is wrong.

              If B = eggs, then B' mean no eggs.

              If one three choices(A, B and C) for breakfast combinations, A, B and C are three different items.
              Last edited by BeanCounter; April 12 2012, 11:29 PM.

              Comment


              • #8
                Originally posted by jchaney View Post
                If one has only three choices for breakfast combination.
                A = Toast
                B = Eggs
                and B' = Ham

                Then we could say that there is a 70% chance of either toast or eggs for breakfast, and 90% chance of toast or ham. Further 30% not toast nor eggs or 10% not toast nor ham. But, what of the those who have toast and eggs, or toast and ham. By this we are saying those had ham are equal to everyone less those who had eggs.

                I get P(a) + p(b) - P(anb) = 1 and P(a) + p(b') - P(anb') = 1
                but, for p(b)+ p(b') = 1 would not p(anb) + p(a) equal to p(b') and consequently p(a) + p(anb') = p(b)? Where is my logic flawed?
                I'm not sure if you're a bad troll or if you are really this unprepared for 1/P...

                Comment


                • #9
                  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.
                  Last edited by jchaney; April 13 2012, 01:41 AM.

                  Comment


                  • #10
                    Originally posted by jchaney View Post
                    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.
                    Prob. For Risk Mgmt 3-47 is directly from SOA sample questions for exam P. Therefore, the notations used in the textbook may be different from notations in those problems.

                    I did not say whether any events in your original question is mutually exlusive. I am referring to one (quoted below) of your examples and you should notice the word, "if", and which of your message is quoted.

                    Originally posted by jchaney View Post
                    I get P(a) + p(b) - P(anb) = 1 and P(a) + p(b') - P(anb') = 1
                    but, for p(b)+ p(b') = 1 would not p(anb) + p(a) equal to p(b') and consequently p(a) + p(anb') = p(b)? Where is my logic flawed?

                    Comment


                    • #11
                      Originally posted by jchaney View Post
                      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.

                      Comment


                      • #12
                        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.

                        Comment


                        • #13
                          Thank you RossMoney34 and NoMoreExams for your persistence and patience. I understand now.

                          Comment

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