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Math Help - Considering Order in Probability

  1. #1
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    Exclamation Considering Order in Probability

    Question;

    4 prizes allocated at random to 24 prize-winners in a lottery. Of the prizes, 4 are cars, 8 are bicycles and 12 are watches. Ann and Ben are two of the prize-winners. Show that the probability that Ann gets a car and Ben gets a bicycle or a watch is 10/69.

    I understand that the answer can be obtained by:
    4/24 * 20/23 = 10/69

    But wouldn't the above solution fail to factor in the event where Ben's prize is first selected? I.e.:
    20/24 * 4/23

    The (flawed?) solution should thus be:
    (4/24 * 20/23) + (20/24 * 4/23) = 20/69 ?

    I get the same answer (20/69) using the selection method:
    (4C1 * 20C1)/ 24C2 = 20/69

    Am I doing something wrong here?
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  2. #2
    Junior Member
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    Nov 2008
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    I think it works out regardless of order

    If I understood your question correctly, you started with:
    4/24 * 20/23 = 10/69

    If you reverse the selection order, I think you get:
    20/24 * 4/23, which is also 10/69, isn't it?

    - Steve J
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  3. #3
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    You got it!

    I understand this seems like it doesn't account for a possibility but as Steve has calculated, it does account for the case. You have the right answer.
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