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Math Help - Help two events

  1. #1
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    Help two events

    Please help

    "The Heavenrich family of five people randomly picks a name of a family member from a hat to decide for whom to buy a present. What is the probability that nobody picks their own name? If someone picks their own name, all the names are returned and everyone picks again. What is the probability that it took exactly two tries for everyone to pick a name of another family member?
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  2. #2
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    Please help

    "The Heavenrich family of five people randomly picks a name of a family member from a hat to decide for whom to buy a present. What is the probability that nobody picks their own name? If someone picks their own name, all the names are returned and everyone picks again. What is the probability that it took exactly two tries for everyone to pick a name of another family member?
    Ah yes. Probability. I think I know how to do the first part.

    Let's give names to the people just for easier reference: Dad, Mom, Kid 1, kid 2, kid 3.

    Let's say Dad picks first.
    There is a 4/5 chance he will NOT pick his name.

    Next mom picks.
    There is a 3/5, because one name that is NOT dad's is gone, she won't pick name.

    Then kid 1 and kid 2 pick.
    There probability is 2/5 and 1/5, respectively.

    NOw Kid 3 picks.
    There is only one name left, here is where I'm not sure about this anymore, that is not his. So 1/5.

    Finally,
    multiply them together because they are dependent events.
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  3. #3
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    Quote Originally Posted by Truthbetold View Post
    Let's give names to the people just for easier reference: Dad, Mom, Kid 1, kid 2, kid 3. Let's say Dad picks first.
    There is a 4/5 chance he will NOT pick his name.
    Next mom picks.
    There is a 3/5, because one name that is NOT dad's is gone, she won't pick name.
    Then kid 1 and kid 2 pick.
    There probability is 2/5 and 1/5, respectively.
    NOw Kid 3 picks.
    There is only one name left, here is where I'm not sure about this anymore, that is not his. So 1/5.
    Finally,
    multiply them together because they are dependent events.
    Hi there Truthbetold
    The correct answer is \frac{{\left( {76} \right)\left( {44} \right)}}{{(120)^2 }}.
    Does that agree with you approach?
    Last edited by Plato; October 20th 2007 at 04:23 PM.
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