# Help! Conditional Probability-Urn Problem

• December 13th 2009, 07:15 PM
Rosalie
Help! Conditional Probability-Urn Problem
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 Im not certain how to apply it in this case. Im really just baffled! Can anyone help me?
• December 13th 2009, 09:25 PM
matheagle
It's just P(A|B).

$P(Y_3|G_1) ={P(Y_3 and G_1)\over P(G_1)}$

$P(G_1) =20/45$... the chance the first is green.

NOW for the numerator you need order.... so hypergeometric formula isn't the best way to do it.
first must be green, second can be anything left over and then the third must be yellow.
Here you need to write down a few sample events.
• December 13th 2009, 09:32 PM
Rosalie
I don't really understand how to use the formula you gave me but, I tried to do the problem using a probability tree, and this is what I got. Is it the same?
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)
P(1st G, 2nd (R,G,Y), 3rd Y)= (52/1419) + (247/4257) + (52/1419) = (13/99)
• December 13th 2009, 09:38 PM
matheagle
That seems to be the correct numerator.
BUT that's just the numerator, you asked for the conditional probability.