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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does the Pr[1]=2/7 mean?

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0I honestly don't know for this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought it might have something to do with probability

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay so k times as likely

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for eeven 2n even number there is k*(2n) of odd numbers

dan815
 one year ago
Best ResponseYou've already chosen the best response.1the probability of 1 is 2/7 okay

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can test your theory too

dan815
 one year ago
Best ResponseYou've already chosen the best response.1another way to look at this is to load a set up

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0Thanks man I understand :D

dan815
 one year ago
Best ResponseYou've already chosen the best response.1we 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1{1,1,1,1,1,1,2, 3,3,3,3,3,3,4,5,5,5,5,5,5,6}

dan815
 one year ago
Best ResponseYou've already chosen the best response.1see if p(1) is really 2/7 now

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0Thanks dan, your the man
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