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Pulsified333

  • one year ago

This question involves a weighted die. For this problem, assume that all the odd numbers are equally likely, all the even numbers are equally likely, the odd numbers are k times as likely as the even numbers, and Pr[1]=2/7. What is the value of k?

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  1. anonymous
    • one year ago
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    what does the Pr[1]=2/7 mean?

  2. Pulsified333
    • one year ago
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    I honestly don't know for this problem

  3. anonymous
    • one year ago
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    I thought it might have something to do with probability

  4. dan815
    • one year ago
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    okay so k times as likely

  5. dan815
    • one year ago
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    for eeven 2n even number there is k*(2n) of odd numbers

  6. Pulsified333
    • one year ago
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    okay

  7. dan815
    • one year ago
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    the probability of 1 is 2/7 okay

  8. Pulsified333
    • one year ago
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    true

  9. dan815
    • one year ago
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    |dw:1441594202967:dw|

  10. dan815
    • one year ago
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    |dw:1441594315632:dw|

  11. dan815
    • one year ago
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    |dw:1441594344511:dw|

  12. Pulsified333
    • one year ago
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    so k=6

  13. dan815
    • one year ago
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    you can test your theory too

  14. dan815
    • one year ago
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    another way to look at this is to load a set up

  15. Pulsified333
    • one year ago
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    Thanks man I understand :D

  16. dan815
    • one year ago
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    we came to the conclusion that the odds are 6 times more likjely another way to say that is to just have 6 times more odd numbers

  17. Pulsified333
    • one year ago
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    It worked

  18. dan815
    • one year ago
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    {1,1,1,1,1,1,2, 3,3,3,3,3,3,4,5,5,5,5,5,5,6}

  19. dan815
    • one year ago
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    see if p(1) is really 2/7 now

  20. dan815
    • one year ago
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    6/((3*6)+3)=6/21=2/7

  21. Pulsified333
    • one year ago
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    Ah

  22. dan815
    • one year ago
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    :)

  23. Pulsified333
    • one year ago
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    Thanks dan, your the man

  24. dan815
    • one year ago
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    you're welcome! *_*

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