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kyanda17

  • one year ago

A bag contains some marbles, each of which is one of four colors (red, white, blue, and green). At least one of the marbles is red. 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?

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  1. dan815
    • one year ago
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    hm

  2. dan815
    • one year ago
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    let r,w,b,g equal the number of red,white,blue green marbles respectively let k = r+w+b+g, k is the total number of marbles (1) 1 marble of each color is chosen r/k * w/(k-1) * b/(k-2)* g/(k-3)= probability that each color is chosen (2) one white, one blue, and two reds are chosen, w/k * b/(k-1)*r/(k-2)*(r-1)/(k-3) you can do these (3) (4)

  3. anonymous
    • one year ago
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    Wait I just get k=4r-6...

  4. kyanda17
    • one year ago
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    see i help u

  5. anonymous
    • one year ago
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    Sorry, but it didn't help me much because I still can't figure it out :(

  6. dan815
    • one year ago
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    hey flyingpie

  7. dan815
    • one year ago
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    @flyingpie you see k has to be an interger from here and u know r is atleast 1

  8. anonymous
    • one year ago
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    So the smallest is 10?

  9. anonymous
    • one year ago
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    @dan815

  10. dan815
    • one year ago
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    can you take me to your question if u have other work up there i can see

  11. anonymous
    • one year ago
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    @dan815 well at least 4 red because of problem statement, so 4(4)-6=10

  12. anonymous
    • one year ago
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    Wait is k=4r-6 right @dan815

  13. dan815
    • one year ago
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    you cant really be sure, i need to see the rest of the work

  14. dan815
    • one year ago
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    you have to make sure that the solutions for white blue and green are also integers

  15. anonymous
    • one year ago
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    4,3,2,1...

  16. anonymous
    • one year ago
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    @dan815

  17. dan815
    • one year ago
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    again lol.. i cant confirm your answer without work xD

  18. dan815
    • one year ago
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    i mean u can just plug those values in and see if it is working out

  19. anonymous
    • one year ago
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    work: the probabilities have same denominator, but the numerators are rwbg=rwb(r-1)=r(r-1)(r-2)b=r(r-1)(r-2)(r-3), so g=r-1, w=r-2, b=r-3, and k=r+w+b+g=4r-6

  20. anonymous
    • one year ago
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    It works but someone said that the answer is 21 from there packet...

  21. dan815
    • one year ago
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    ya there are infinite answers only 1 lowest

  22. dan815
    • one year ago
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    you can scale all the values up so that the probabilities still work out so

  23. anonymous
    • one year ago
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    What do you think answer is @dan815

  24. dan815
    • one year ago
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    well im just going off by your statements, if u say u check it then 10 is the lowest i guess

  25. dan815
    • one year ago
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    (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. k=r+w+g+b rwgb = wbr*(r-1) = b*r*(r-1)*(r-2)=r*(r-1)(r-2)(r-3)

  26. dan815
    • one year ago
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    http://prntscr.com/88hp94 looks like it is right

  27. anonymous
    • one year ago
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    but could you tell me what is smallest possible? is it 10? Thanks!

  28. anonymous
    • one year ago
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    @dan815

  29. anonymous
    • one year ago
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    hello?

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