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Math Help - Dice Problem

  1. #1
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    Dice Problem

    B throws 12 dice and wins if he scores at least two aces.
    Find the probability:

    So I was wondering if anyone could confirm or deny that
    P(A) = 1 - (5/6)^12 - ((5/6)^11)(1/6)

    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    B throws 12 dice and wins if he scores at least two aces.
    Find the probability:

    So I was wondering if anyone could confirm or deny that
    P(A) = 1 - (5/6)^12 - ((5/6)^11)(1/6)

    Thanks
    It should be:  1 - \left( \frac56 \right)^{12}  - {{12}\choose 1}\left( \frac56 \right)^{11} \left( \frac16\right)
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    i'm confused

    what does "aces" mean when we're talking about dice?
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  4. #4
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    It means rolling a one.
    But why do you need to multiply by 12 in the expression?
    Thanks
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  5. #5
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    Quote Originally Posted by tbyou87 View Post
    It means rolling a one.
    But why do you need to multiply by 12 in the expression?
    Thanks
    Because we are thinking of this probably as an ordered pair:
    X X X X X X X X X X X X
    Now you want the probably of getting exactly one 1.
    There are twelve's locations for that 1. And each have (5/6)^11 (1/6) probability.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    It means rolling a one.
    But why do you need to multiply by 12 in the expression?
    Thanks
    i hope you noticed that he used a binomial distribution method here. it might not have been what he thought about, but it's what he ended up using. the expression he has is 1 - Bin(12,0) - Bin(12,1), which is the probability of NOT rolling 0 aces or one ace, that is, at least two
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