Urn Conditional Probability Problem

AI Thread Summary
The discussion revolves around calculating the probability of drawing a yellow ball third, given that the first ball drawn is green, from an urn containing 12 red, 20 green, and 13 yellow balls. Participants explore using Bayes' theorem and probability tree diagrams to determine the correct probabilities for various combinations of draws. Initial calculations reveal the probabilities for different sequences involving green and yellow balls, leading to a total probability of 13/99 for the specific scenario. A key insight is that the probability of drawing a yellow ball is consistent regardless of the second ball drawn, simplifying the calculations. The conversation emphasizes the importance of correctly applying Bayes' theorem to arrive at the conditional probability.
Rosalie
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The problem is stated as follows:

You have an urn with 12 red balls, 20 green balls, and 13 yellow balls. Suppose 3 balls are drawn without replacement. What is the probability the third ball is yellow given the first ball is green.

I am pretty sure I use Bayes formula, but I am not certain how to apply it in this case. I am really just baffled! Can anyone help me?
 
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First step is to work out the probability of each combination of 3 balls. Might be easiest with the aid of a probability tree diagram.
 
Im not certain I understand how to set up a probability tree for a problem like this. I am going to try to work it through. Give me a moment, I'll let you double check my results.
 
Ok! I think I got it. This is what I concluded.
Total Balls=45
Green(G)=20
Red(R)=12
Yellow(Y)

P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

If this is incorrect please let me know! It is very important I get this question correct. Thank you for all your help.
 
Rosalie said:
P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

You're half-way there - what you've worked out is the probability that 3rd is yellow AND 1st is green. Next step uses Bayes formula.
 
Rosalie said:
Ok! I think I got it. This is what I concluded.
Total Balls=45
Green(G)=20
Red(R)=12
Yellow(Y)

P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

If this is incorrect please let me know! It is very important I get this question correct. Thank you for all your help.

I can't resist. Did you notice that the value for P(G,Y) is also 13/99?

This is no accident. As long as you have no information about the second ball, the third ball and the second ball have exactly the same probability distributions. An easy way to get this first step would be to observe this equivalence and then just calculate P(G,Y) = (20/45)*(13/44)

It all makes no difference to the next step, where you apply Bayes theorem to get the required conditional probability.

Cheers -- sylas
 
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