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Math Help - Probability--Rolling a die

  1. #1
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    Probability--Rolling a die

    This probably should be very easy but i am not good at probability.

    Given a fair die, what is the probability of rolling at least 2 sixes if the die is rolled 12 times.

    I have tried a little bit, and this is what I have.

    First of all, i know the universe is equal to 6^12. And the hint is that it is easier to compute the probability of the event not happening.
    If A is the event of getting at least 2 sixes, B is getting 1 or no sixes.

    P(B)=?? Therefore, P(A)=1-P(B)


    Thanks for any help.
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  2. #2
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    Hello, dude15129!

    You say you're not good at probability,
    . . but you have a good understanding of it.

    Your game plan is absolutely correct!


    Number of ways to get no 6's: . \left(\frac{5}{6}\right)^{12}

    Number of ways to get one 6: . {12\choose1}\left(\frac{1}{6}\right)\left(\frac{5}  {6}\right)^{11}


    Hence: . P(\text{no 6 or one 6}) \:=\:\frac{\left(\frac{5}{6}\right)^2 + 12\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)  ^{11}} {6^{12}} \;=\;\frac{830,\!078,\!125}{2,\!176,\!782,\!336}


    Therefore: . P(\text{two or more 6s}) \;\;=\;\;1 - \frac{830,\!078,\!125}{2,\!176,\!782,\!336} \;\;=\;\;\frac{1,\!346,\!704,\!211}{2,\!176,\!782,  \!336}

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  3. #3
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    Thanks for your help,
    but can you explain the way you got the number of ways to get 1 six? i don't understand the formula/(12 1) thing.


    Also, the next part asks about the probability of rolling at least 3 sixes if the die is rolled 18 times. So, following what you did.... it would be....


    P(rolling 0,1,2 sixes)= [(5/6)^18 + 18(1/6)(5/6)^17 + 18(1/6)(5/6)^16]/6^18


    Thanks.
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  4. #4
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    Quote Originally Posted by dude15129 View Post
    Also, the next part asks about the probability of rolling at least 3 sixes if the die is rolled 18 times. So, following what you did.... it would be....
    [FONT=monospace]P(rolling 0,1,2 sixes)= [(5/6)^18 + 18(1/6)(5/6)^17 + 18(1/6)(5/6)^16]/6^
    Are you sure that this is not fully explained in your text material?

    If we roll a die 18 times, the the probability of getting exactly k\text{ sixes },~0\le k \le 18 is given using the binominal formula.
    \binom{18}{k}\left(\frac{1}{6}\right)^k\left(\frac  {5}{6}\right)^{18-k}

    That must be somewhere in your notes. If not, then you are not ready for this problem.
    Last edited by Plato; August 29th 2009 at 03:16 PM.
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  5. #5
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    haha....well actually we haven't had any notes really and our textbook isn't like a real textbook.

    but thanks for the help.
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