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anonymous

  • one year ago

Will Medal and Fan: Hard Probability Question (need thorough explanation): A bag contains some marbles, each of which is one of four colors (red, white, blue, and green). There is at least one of each color. The composition of the bag is such that if we take four marbles out at random (without replacement), each of the following is equally likely: (1) one marble of each color is chosen, (2) one white, one blue, and two reds are chosen, (3) one blue and three reds are chosen, (4) four reds are chosen. What is the smallest possible number of marbles in the bag? Thanks.

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  1. anonymous
    • one year ago
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    one or two

  2. dan815
    • one year ago
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    supposed you have r number of red w number of white b numb of blue and g number of green

  3. dan815
    • one year ago
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    the prob you pick out r w b g in that order is r/(w+b+g) * w/((w+b+g)-1) * b/((w+b+g)-2) * g/((w+b+g)-3) now this is only one case where we get it in that order, ther are 4! ways where we still get r w b g in the end so the total prob of getting one of each color is 4!*(r/(w+b+g) * w/((w+b+g)-1) * b/((w+b+g)-2) * g/((w+b+g)-3) )

  4. dan815
    • one year ago
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    you can repeat this process, for all the 4 statments, and you will end up with 4 equations and 4 unknowns

  5. dan815
    • one year ago
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    The thing here is, that the solution you get will be the least number of each color needed, this will hold true if we scale all the balls up by the same factor, as the proportions will be the same,

  6. dan815
    • one year ago
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    |dw:1435348899747:dw|

  7. dan815
    • one year ago
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    |dw:1435349016423:dw|

  8. dan815
    • one year ago
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    |dw:1435349098287:dw|

  9. dan815
    • one year ago
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    so all of them = K and see the relationship we get for rwgb

  10. dan815
    • one year ago
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    we can continue when you are back

  11. kropot72
    • one year ago
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    Let the numbers of marbles be r(red), w(white), b(blue) and g(green). Let the total number of marbles be s = r + w + b + g \[\large P(1r,1w,1b,1g)=\frac{rC1\times wC1\times bC1\times gC1}{sC4}\ ..........(1)\] \[\large P(1w, 1b, 2r)=\frac{rC2\times wC1\times bC1}{sC4}\ .............(2)\] \[\large P(1b, 3r)=\frac{rC3\times bC1}{sC4}\ .........(3)\] \[\large P(4r)=\frac{rC4}{sC4}\ ..........(4)\] The four values of probability are given as being equal. The denominators of (1), (2), (3) and (4) are equal, therefore the numerators must be equal. Equating the numerators of (1) and (2) and simplifying gives: \[\large rC2=rC1\times gC1=r \times g\ ............(5)\] From equation (5) it can be found that possible combinations of r and g are r = 5, g = 2 ; r = 7, g = 3 ; r = 9, g = 4 ; r = 11, g = 5 ; r = 13, g = 6 ; and so on. The next step is to find the values of rC4 for r = 5, 7, 9, 11, 13. By substitution of each of the five possible combinations of r and g shown above into the numerators of (1), (2), (3) and (4), the lowest possible product of w and b is found to be 6, when r = 11 and g = 5. From there it is simple to find the values of w and b.

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