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# Bag and balls revisited

11 replies to this topic

### #1 bonanova

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Posted 29 July 2012 - 12:32 AM

A bag contains two blue balls and an undetermined number of red balls.
A ball of unknown (red or blue) color is added to the bag.
Finally, a ball is drawn at random from the bag.

If the drawn ball is blue, what is the probability that the added ball was red?

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### #2 phil1882

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Posted 29 July 2012 - 02:15 AM

i would be very curious to see the logical answer to this... my guess would be...
Spoiler for

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### #3 bushindo

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Posted 29 July 2012 - 07:08 AM

A bag contains two blue balls and an undetermined number of red balls.
A ball of unknown (red or blue) color is added to the bag.
Finally, a ball is drawn at random from the bag.

If the drawn ball is blue, what is the probability that the added ball was red?

I think the crux of the puzzle lies in the part that is highlighted red above. I assume that by 'undetermined number of red balls', you mean that the precise number of red balls is unknown. However, the solution to the puzzle requires that we assume something (a priori information) about the generating distribution of the red balls (e.g., poisson distribution, exponential distribution, gamma distribution, etc.). We could also start going into improper priors, but I'd rather not =). This puzzle seems dangerously close to a subjective argument about what type of prior distribution does 'undetermined number of red balls' mean, maybe the OP can clarify that a bit?
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### #4 benjer3

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Posted 29 July 2012 - 08:52 AM

Spoiler for My Thought Process

Edited by benjer3, 29 July 2012 - 08:52 AM.

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### #5 kbrdsk

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Posted 29 July 2012 - 11:48 AM

Spoiler for guess

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### #6 bushindo

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Posted 29 July 2012 - 12:25 PM

I was incorrect in the post above. Turns out we don't really need the prior distribution of the red balls. In the words of CaptainEd, "Durn, hit by Bayes again" :-)
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### #7 jim

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Posted 29 July 2012 - 01:57 PM

EXTENSION PROBLEM--only slightly more difficult. Instead of assuming the added ball had an original probabilty of 1/2 red and 1/2 blue assume a probability of r red and b blue, with r+b = 1 of course.
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### #8 plasmid

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Posted 29 July 2012 - 04:00 PM

Spoiler for the extended form of the problem

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### #9 kbrdsk

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Posted 29 July 2012 - 04:04 PM

Spoiler for another guess

Edited by kbrdsk, 29 July 2012 - 04:04 PM.

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### #10 jim

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Posted 29 July 2012 - 09:30 PM

EXTENSION 2--Same problem but now assume we add two balls red with probability r and blue with probability b. What about when we add three balls? This time it will we convienent to assume we start with 10 blue balls--but it is easy to generalize to k blue balls, just a little messy to write up the answer.
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