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Thread: Set Theory: find P(A)

  1. #1
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    Set Theory: find P(A)

    P(AuB) = .7
    P(AuB')=.9

    Find P(A)

    So we know everything outside of B is .9
    We know A or B =.7

    We want to find A.

    I am kind of stuck here.


    I think I need to know the outside

    of A and B, which is P(AuB)' which is = A' intersect B'
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  2. #2
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    Re: Set Theory: find P(A)

    Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.
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  3. #3
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    Re: Set Theory: find P(A)

    Quote Originally Posted by HallsofIvy View Post
    Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.
    I think you mean AnB and AnB' are disjoint?? And P(A)=P(AnB)+P(AnB') ??
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  4. #4
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    Re: Set Theory: find P(A)

    Set Theory: find P(A)-venn-diagram.jpg
    I find these easier to do using a Venn diagram. See attachment.
    Thanks from math951
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  5. #5
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    Re: Set Theory: find P(A)

    Quote Originally Posted by math951 View Post
    P(AuB) = .7
    P(AuB')=.9
    Find P(A)
    While I agree that the use of Venn diagrams, see reply #4, are useful, here is another approach.
    We know that $\mathscr{P}(A)=\mathscr{P}(A\cap B)+\mathscr{P}(A\cap B')$
    $ \begin{align*}\mathscr{P}(A\cup B)&=\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A\cap B) \\\mathscr{P}(A\cup B')&=\mathscr{P}(A)+\mathscr{P}(B')-\mathscr{P}(A\cap B')\\\mathscr{P}(A\cup B)+\mathscr{P}(A\cup B')&=2\mathscr{P}(A)+\mathscr{P}(B)+( B')-\mathscr{P}(A\cap B)-\mathscr{P}(A\cap B')\\0.7+0.9&=2\mathscr{P}(A)+1-\mathscr{P}(A)\\\mathscr{P}(A)&=0.6 \end{align*}$
    Thanks from SlipEternal and math951
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