A little help understanding this bayesian problem (very basic)

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SUMMARY

This discussion centers on a Bayesian probability problem involving prior probabilities for red balls in an urn. The user initially calculated the prior probabilities as 1/10, assuming 10 possible outcomes (0 to 9 red balls). However, the book's answer suggests a prior of 1/9, leading to confusion about the correct interpretation of the problem. The consensus among participants indicates that either the zero balls case should be excluded or all priors should indeed be 1/10, confirming the user's calculations as valid.

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Homework Statement


so I'm trying to teach myself bayes, and i got a book, and I'm going through trying to do the exercises, and lo and behold, i get stuck on the first one. i thought i was getting it, but the answer given at the back of the book is different than mine.



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The Attempt at a Solution


i've attached three pictures. the first is the actual problem. so when i did it myself, i got that the first column of priors should all be 1/10, since we're assuming that the urn can contain 0 up to 9 red balls, for a total of 10 possibilities. the second picture is the answer, where it seems that i should have said 1/9 for the first column. the third picture is an example that seems to me to indicate that it should be 1/10. what am i missing?
 

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There are 10 possible values for X, but there are only 9 balls to choose from.
What does the prior probability represent in this case?
 
the prior probability represents the number of red balls in the urn. I'm told to assume that each possibility is equally likely. so there might be zero balls, 1 ball, 2 balls, etc, up to 9 balls. that should be a 1/10 probability for each case. furthermore, shouldn't the prior column sum up to 1? in the answer, doesn't the prior column sum up to 10/9?
 
bennyska said:
the prior probability represents the number of red balls in the urn. I'm told to assume that each possibility is equally likely. so there might be zero balls, 1 ball, 2 balls, etc, up to 9 balls. that should be a 1/10 probability for each case. furthermore, shouldn't the prior column sum up to 1? in the answer, doesn't the prior column sum up to 10/9?

You are correct. Either the "zero balls" case should not be present (giving 1/9 probability of each of the others) or else it is present and all priors should be 1/10.

RGV
 
thank you. also, i should have been more clear, the third picture was an example with the exact same problem, just fewer balls.
 
Yes - I'm inclined to agree that the text is mistaken in the model answer.
Frustrating, I know.

I had a go trying to identify the mistake but failed. Why would the author think to assume that P(X=x)=x/(N-1)?

I'd ignore the answers but do the analysis anyway. Pause at each step to see if what you get makes sense. The other one seems to be right - maybe you want to start with that instead.
 

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