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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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
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what does the Pr[1]=2/7 mean?
I honestly don't know for this problem
I thought it might have something to do with probability

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Other answers:

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

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