Have You Seen This Urn Problem Before?

  • Thread starter Thread starter pieface
  • Start date Start date
AI Thread Summary
The urn problem involves drawing pairs of colored balls (red, green, blue, and orange) from an urn using a sampling machine that only draws two colors at a time. The challenge is to estimate the proportion of each color in the urn based on the outcomes of these pairs. The discussion clarifies that the balls are drawn with replacement, and there are six possible outcomes for the pairs. To estimate the proportion of a specific color, the number of pairs containing that color is divided by twice the total number of drawn pairs. This problem does not have a specific name but falls under various urn problems in probability theory.
pieface
Messages
2
Reaction score
0
I'm interested if anyone has seen something like this urn problem before and if it has a name?

The premise is:

An urn contains red, green, blue and orange balls. I have a sampling machine that can draw samples from the urn, but will only draw balls of any two colors at a time. For example, the machine can draw red and green balls and I will get a sample that tells me the ratio of red to green but not the proportion of green balls in the urn as a whole.

Thanks!
 
Physics news on Phys.org
I'm assuming your problem is about drawing a pair of two different color balls each time, right? Do you replace the balls back to the urn before the next drawing?
 
Hi Viralux, yes, different colored balls, and with replacement. Thanks
 
I don't think this problem has an specific name since there are many different urn like problems in the probability theory.

In this case you want to estimate the proportion of each colored ball; if you draw one at a time you simply divide the number of of each colored ball n by the total number of drawn balls N, that is n/N.

Since you are drawing two at a time and they are different color each time you have only 6 possible outcomes {r,g} {r,b} {r,o} {g,b} {g,o} {b,o} and you can estimate then the proportion for each pair as if they were just one color.

But if you want to estimate the proportion of just one of the colors you simply need to divide the number of pairs having that color by twice the total number of drawn pairs, for example, the estimation for the red ball would be: (#{r,g}+#{r,b}+#{r,p})/(2*N)
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top