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Math Help - Probability qn 2

  1. #1
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    Probability qn 2

    Hi I really need help with this too! I cannot see where I went wrong, really. Thank you so much in advance! Really need this!

    ! I've attached the question itself and my attempt to find P(D) to which I failed.


    .
    Attached Thumbnails Attached Thumbnails Probability qn 2-math-qn.jpg   Probability qn 2-math-question.jpg  
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  2. #2
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    Re: Probability qn 2

    It is hard to read your images.
     \begin{align*} \mathcal{P}(C) &= \mathcal{P}(C\cap D)+ \mathcal{P}(C\cap D') \\\frac{9}{20}&=\frac{6}{13}\mathcal{P}(D)+\frac{3  }{7}(1-\mathcal{P}(D))\end{align*}

    Solve for \mathcal{P}(D).
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  3. #3
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    Re: Probability qn 2

    Thank you so much!
    I didn't know that P(C|D)+P(C|D') = P(C)!
    If I expand it out eg. P(C|D)P(D) = P(CnD), it will be quite different from P(C|D')P(D') = P(CnD') so I dont think that's the proof..

    Could you teach me how to prove P(C|D)+P(C|D') = P(C) using any method or even the venn diagram?
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  4. #4
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    Re: Probability qn 2

    Quote Originally Posted by Tutu View Post
    I didn't know that P(C|D)+P(C|D') = P(C)!
    If I expand it out eg. P(C|D)P(D) = P(CnD), it will be quite different from P(C|D')P(D') = P(CnD') so I dont think that's the proof..
    That is not true. Nor did I claim it was.

    But it is true that
    \mathcal{P}(C)=\mathcal{P}(C\cap D)+\mathcal{P}(C\cap D')=\mathcal{P}(C|D)\mathcal{P}(D)+\mathcal{P}(C|D  ')\mathcal{P}(D').
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  5. #5
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    Re: Probability qn 2

    I understand with the help of venn diagrams! Thank you so much!
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