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Math Help - Conditional probability

  1. #1
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    Conditional probability

    Imagine you have a wall with a hole.Behind the wall there are 10 targets.On the other side, you have a box with 52 darts.Only one of the darts has the word throw on it.Each target will be showed in the hole in sequence.On every target showed you will pick a dart.If the dart has the word throw, youŽll throw the dart and another dart with the word throw will be put in the box, if not youŽll put it back in the box and youŽll repeat the operation.

    if a dart is thrown what is the probability of the target #5 be in the whole ?
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  2. #2
    o_O
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    A = #5 target showing up in the hole
    B = Dart labelled 'throw' is picked

    By definition:
    P(A \big| B) = \frac{P(A \cap B)}{P(B)}

    P(A \cap B) = {\color{red}\frac{1}{10}} \cdot {\color{blue}\frac{1}{52}} \qquad P(B) = {\color{blue}\frac{1}{52}}

    Only one target out of 10 is labelled #5. Only 1 dart out of 52 is labelled 'thrown'.

    ---

    Or if you simply thought about the problem. Since you're given the fact that a dart is thrown, all you have to consider is the probability of hitting the #5 target out of 10 possible targets.
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  3. #3
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    Quote Originally Posted by o_O View Post
    A = #5 target showing up in the hole
    B = Dart labelled 'throw' is picked

    By definition:
    P(A \big| B) = \frac{P(A \cap B)}{P(B)}

    P(A \cap B) = {\color{red}\frac{1}{10}} \cdot {\color{blue}\frac{1}{52}} \qquad P(B) = {\color{blue}\frac{1}{52}}

    Only one target out of 10 is labelled #5. Only 1 dart out of 52 is labelled 'thrown'.

    ---

    Or if you simply thought about the problem. Since you're given the fact that a dart is thrown, all you have to consider is the probability of hitting the #5 target out of 10 possible targets.
    If instead of I say that only one dart is thrown , I say that at least one dart is thrown in the 10 attempts , does the calculation change ?
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