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Math Help - Two events and a simple question

  1. #1
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    Two events and a simple question

    what the question states :
    "Events A and B are such that:
    p(A) = 5/12
    p(A|notB)=7/12
    P(A intersection B ) =1/8 "

    Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.
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  2. #2
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    Re: Two events and a simple question

    Quote Originally Posted by terence View Post
    what the question states :
    "Events A and B are such that:
    p(A) = 5/12
    p(A|notB)=7/12
    P(A intersection B ) =1/8 "

    Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.
    no you cant assume that, not B not always just A, if A union B is the whole space then not B is in A
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  3. #3
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    Re: Two events and a simple question

    Most likely "notB" means "event B did not occur"

    I would also advise you to draw a Venn diagram for this question.
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  4. #4
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    Re: Two events and a simple question

    Quote Originally Posted by terence View Post
    what the question states :
    "Events A and B are such that:
    p(A) = 5/12
    p(A|notB)=7/12
    P(A intersection B ) =1/8 "
    Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.
    P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}.
    Now P(B^c)=1-P(B).
    Thus P(A\cap B^c)=P(A|B^c)P(B^c)
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  5. #5
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    Re: Two events and a simple question

    Quote Originally Posted by Plato View Post
    P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}.
    Now P(B^c)=1-P(B).
    Thus P(A\cap B^c)=P(A|B^c)P(B^c)
    The question then asks for :

    i) P(B)
    ii) P(A|B)
    iii) P(B|A)
    iv) P(A U B)


    P(B|A) is easy since P(A intersection B) / P(A) = 0.3
    Had i known P(B) iv) would have been easy as well.

    How can P(B) be found.. ?
    Well I know this:
    \begin{align*}P(B) &= P(A\cap B) &+ P(notA\cap B)\end{align*}
    where
    \begin{align*}P(notA\cap B) &=P(A).P(B|notA)\end{align*}
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  6. #6
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    Re: Two events and a simple question

    Quote Originally Posted by terence View Post
    The question then asks for :
    i) P(B)
    ii) P(A|B)
    iii) P(B|A)
    iv) P(A U B)
    How can P(B) be found.. ?
    Well I know this:
    \begin{align*}P(B) &= P(A\cap B) &+ P(notA\cap B)\end{align*}
    where
    \begin{align*}P(notA\cap B) &=P(A).P(B|notA)\end{align*}
    From the given you can find P(A\cap B^c).
    But P(A\cap B^c)=P(A|B^c)P(B^c) which is given.
    From which you can find P(B^c). From which you can find P(B).
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  7. #7
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    Re: Two events and a simple question

    From the given you can find P(A\cap B^C)
    how can you find this ?

    the nearest i can get is that
    P(A\cap B^C)  &= 7/12 * P(B)
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  8. #8
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    Re: Two events and a simple question

    Quote Originally Posted by terence View Post
    how can you find this ?
    P(A\cap B^c)=P(A)-P(A\cap B)
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  9. #9
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    Re: Two events and a simple question

    Oh how dumb of me -.-

    I should have reasoned that out. Largest problem with probability=> reasoning simple things out.


    Thanks @Plato
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