Question 1180491: 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[6]=1/18.
What is the value of k?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
The problem as stated is nonsense and can't be solved. There are an infinite number of even numbers and an infinite number of odd numbers; it makes no sense to say P(6)=1/18.
If this is a problem about a standard fair six-sided die, then you need to say so.
In any case, you need to define the complete finite set of numbers that the problem is about.
Re-post defining the problem completely.
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The reader posted a thank-you note saying he had solved the problem and the answer is k=5.
Tutor @ikleyn posted a response saying k=17.
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The response from tutor @ikleyn is absurd. If k=17 and P(6)=1/18, then the probability of any odd number would be 17/18. Since 1/18+17/18=1, the conclusion from that would be that n is some odd number and the universal set the problem is working with consists of two elements -- 6 and some odd number.
The statement of the problem says nothing about the set of numbers the problem is about; but a set of "6" and some odd number is unlikely.
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What the reader did not say is that the problem is about rolling a standard 6-sided die.
In that case, given P(6)=1/18, we know that P(2)=1/18 and P(4)=1/18, so P(even)=3/18.
That means P(odd)=15/18; and since all odd numbers are equally likely, P(1)=P(3)=P(5)=5/18.
So each odd number has a probability that is 5 times the probability of an even number.
So k=5.
Answer by ikleyn(52852)  (Show Source):
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