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Math Help - probability question-marbles.

  1. #1
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    probability question-marbles.

    there are a certain number of marbles in a bag, some are green, some red and some blue. I pick 3 blue marbles in a row. i observe that there was a 1 in 3417 chance that this would occur. if only my first two marbles where blue there would be a 1 in 204 chance.

    -what is the smallest number of marbles that can be in my bag?
    -how many blue marbles are there?
    -how many possibllities are there for numbers of marbles in my bag?

    i really need any help i can get with this, been working on it for hours
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  2. #2
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    Let the number of blue marbles be b.
    Let the total number of marbles be n.
    P(3 blue marbles in a row) =b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417
    P(2 blue marbles in a row) =b/n * (b-1)/(n-1) = 1/204

    If we replace the second relation into the first one, we get:


    1/204 * (b-2)/(n-2) = 1/3417

    Hence: 3417 * (b-2) = 204 * (n-2)

    Finding the least number of marbles in the bag boils down to finding the LCM(204, 3417) = 2 * 2 * 3 * 17 * 67 = 13668

    So: 3417 * 4 = 204 * 67

    One possible solution: b-2 = 4 and n-2 = 67

    b=6 and n= 69. When replaced back in the initial condition b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417 it does not work

    Therefore we have to go for multiples of 4 and 67

    The next possible solution: b-2 = 2*4 and n-2 = 2*67
    b=10 and n=136 - which is the correct one.

    For your third question, just try to play woth multiples of 4 and 67.
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  3. #3
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    in response

    Quote Originally Posted by Ithaka View Post
    Let the number of blue marbles be b.
    Let the total number of marbles be n.
    P(3 blue marbles in a row) =b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417
    P(2 blue marbles in a row) =b/n * (b-1)/(n-1) = 1/204

    If we replace the second relation into the first one, we get:


    1/204 * (b-2)/(n-2) = 1/3417

    Hence: 3417 * (b-2) = 204 * (n-2)

    Finding the least number of marbles in the bag boils down to finding the LCM(204, 3417) = 2 * 2 * 3 * 17 * 67 = 13668

    So: 3417 * 4 = 204 * 67

    One possible solution: b-2 = 4 and n-2 = 67

    b=6 and n= 69. When replaced back in the initial condition b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417 it does not work

    Therefore we have to go for multiples of 4 and 67

    The next possible solution: b-2 = 2*4 and n-2 = 2*67
    b=10 and n=136 - which is the correct one.

    For your third question, just try to play woth multiples of 4 and 67.
    Thanks very much, i found out that by simultaneosly solving (b-2)/4=(n-2)/67 and b/n * (b-1)/(n-1) = 1/204 you get the quadratic equation (for b):

    175b^2 - 22207b - 216648 = 0

    the other solution is -1593/175.
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