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Thread: set operations noob question

  1. #1
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    set operations noob question

    I tried solving some of the problems for the SOA 150 practice questions and there was something I just couldnt get.

    I'm studying from the finan free e-book and other statistic specific resources.

    the problem states. having P[AUB]=0.7 & P[AUB']=0.9 find P[A].

    so I expand both and add them together and get

    1.6= 2P[A] - (P[B] - P[B']) + P[(ANB)U(ANB')] <-last term is equal to P[A] because they are disjoint, I got that, but I don't know what to do with
    (P[B] - P[B']).

    according to the SOA (P[B] - P[B'])= 1 but why? I cant find any rule that says that.

  2. #2
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    Are you sure that's what it says?

  3. #3
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    yes, just rechecked it

  4. #4
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    Quote Originally Posted by mefisto66 View Post
    yes, just rechecked it
    What problem is that?

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  6. #6
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    Don't they say P(B) + P(B') = 1.... you said P(B) - P(B') = 1

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    oops, my bad. but still, what rule is that?

  8. #8
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    Quote Originally Posted by mefisto66 View Post
    oops, my bad. but still, what rule is that?
    Not sure it has a name but doesn't it make sense? P(A) is everything in A, P(A') is everything not in A. The sum of those 2 should be everything or since it's probability... 1

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    Dear Add up the both

    P(AUB)=.7, P(AUB')=.9

    1.6=P(A)+P(B)+P(A)+P(B')-{AnB)+(ANB')}
    1.6=2P(A)+{P(B)+[1-P(B)-P(A)

    PLEASE NOTE:- IF [B+1-B] WHICH CALCELLES THE B & -B AND REMAINDER IS 1, THEREFORE [P(b)+(b')]=1

    1.6=P(A)+1
    1.6-1=P(A)
    0.6=P(A)

    Think, now you would be clear
    Last edited by imrancairo; September 21st 2011 at 08:27 AM.

  10. #10
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    Quote Originally Posted by imrancairo View Post
    Dear Add up the both

    P(AUB)=.7, P(AUB')=.9

    1.6=P(A)+P(B)+P(A)+P(B')-{AnB)+(ANB')}
    1.6=2P(A)+{P(B)+[1-P(B)-P(A)

    PLEASE NOTE:- IF [B+1-B] WHICH CALCELLES THE B & -B AND REMAINDER IS 1, THEREFORE [P(b)+(b')]=1

    1.6=P(A)+1
    1.6-1=P(A)
    0.6=P(A)

    Think, now you would be clear
    He was asking WHY does P(B) + P(B') = 1 which I think I answered.

    Additionally your statement of "IF [B+1-B] WHICH CALCELLES THE B & -B AND REMAINDER IS 1, THEREFORE [P(b)+(b')]=1" doesn't really make much sense, your antecedent is an expression and your consequent is an equality.

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