A little help understanding this bayesian problem (very basic)

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Homework Help Overview

The discussion revolves around a Bayesian probability problem involving prior probabilities for the number of red balls in an urn. The original poster is attempting to understand the correct assignment of prior probabilities based on the problem's setup and the provided answer in their textbook.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the assignment of prior probabilities, questioning whether the prior should be 1/10 or 1/9 based on the number of possible outcomes. There is discussion about the implications of including the case of zero balls and the requirement for the prior probabilities to sum to 1.

Discussion Status

Participants are actively questioning the assumptions made in the problem and the textbook's answer. Some express frustration over potential errors in the model answer, while others suggest analyzing the problem step by step to ensure understanding.

Contextual Notes

There is a mention of a specific example with fewer balls that may provide additional context, but the relevance of this example to the current problem is not fully explored.

bennyska
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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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  • bayes - answers.jpg
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  • bayes - prob1.jpg
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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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