Recent content by Rosalie

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...

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.

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...