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Math Help - Conditional Probability

  1. #1
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    Conditional Probability

    I'm having some serious difficulty with the question on XP 63a



    I get (a).

    For (b), I don't get how the first line is derived.

    I also don't know why p(B|\bar A)=\frac{3}{9}

    For (c), I don't get anything, how it is derived, or how it is calculated.
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  2. #2
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    Quote Originally Posted by chengbin View Post
    I'm having some serious difficulty with the question on XP 63a I get (a).
    For (b), I don't get how the first line is derived.
    I also don't know why p(B|\bar A)=\frac{3}{9}
    For (c), I don't get anything, how it is derived, or how it is calculated.
    I do not think that any of us can follow your question.
    The a, b, c do not refer to anything in that picture.
    So instead of posting a copy of a page, why not simply type out your question?
    Then we will be sure of what you are asking.
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  3. #3
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    Its a pain to type so much word and code.

    My question is on the right side of the page (XP 63a)

    There is a question, and there are 3 ways of answering it (case a, b, and c)

    (a), that is obvious.

    (b) How do you get the first line? ( B=(A\cap B)\cup (\bar A \cup B))

    (c) Why does C equal to that? (Too lazy to type that much code, sorry)
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  4. #4
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    Quote Originally Posted by chengbin View Post
    Its a pain to type so much word and code.
    (Too lazy to type that much code, sorry)
    Well that is too bad. I for one don't know why we should answer.
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  5. #5
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    OK, OK (although I still don't see why I should type it out when I provided a very clear image of my question)

    Why is C=(A\cap B\cap C)\cup (\bar A \cap B \cap C)\cup (A\cap \bar B \cap C)\cup (\bar A \cap \bar B \cap C)

    Whoo! That was quite a bit of work. It was difficult with all these similarities!
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  6. #6
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    Quote Originally Posted by chengbin View Post
    Why is C=(A\cap B\cap C)\cup (\bar A \cap B \cap C)\cup (A\cap \bar B \cap C)\cup (\bar A \cap \bar B \cap C)
    Because (A\cap B\cap C)\cup (\bar A \cap B \cap C)\cup (A\cap \bar B \cap C)\cup (\bar A \cap \bar B \cap C) there we have four pair-wise disjoint sets.
    Consider any element of C. Then that element will belong to one of those four sets.
    The importance of being pair-wise disjoint sets is that the probability of the union is simply the sum of the four probabilites.

    If you know how to do Venn digrams, then it is easy to see.
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