Pulsified333
  • Pulsified333
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
what does the Pr[1]=2/7 mean?
Pulsified333
  • Pulsified333
I honestly don't know for this problem
anonymous
  • anonymous
I thought it might have something to do with probability

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More answers

dan815
  • dan815
okay so k times as likely
dan815
  • dan815
for eeven 2n even number there is k*(2n) of odd numbers
Pulsified333
  • Pulsified333
okay
dan815
  • dan815
the probability of 1 is 2/7 okay
Pulsified333
  • Pulsified333
true
dan815
  • dan815
|dw:1441594202967:dw|
dan815
  • dan815
|dw:1441594315632:dw|
dan815
  • dan815
|dw:1441594344511:dw|
Pulsified333
  • Pulsified333
so k=6
dan815
  • dan815
you can test your theory too
dan815
  • dan815
another way to look at this is to load a set up
Pulsified333
  • Pulsified333
Thanks man I understand :D
dan815
  • dan815
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
Pulsified333
  • Pulsified333
It worked
dan815
  • dan815
{1,1,1,1,1,1,2, 3,3,3,3,3,3,4,5,5,5,5,5,5,6}
dan815
  • dan815
see if p(1) is really 2/7 now
dan815
  • dan815
6/((3*6)+3)=6/21=2/7
Pulsified333
  • Pulsified333
Ah
dan815
  • dan815
:)
Pulsified333
  • Pulsified333
Thanks dan, your the man
dan815
  • dan815
you're welcome! *_*

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