the question is pretty long so I didn't write it in the title.
The Question:
there are 3 urns marked A, B and C. In urn A there are 2 white balls and 4 red balls. In urn B there are 8 white balls and 4 red balls. In urn C there is one white ball and three red balls. We choose one ball from each of the urns (uniformly at random from each urn and independently between the different urns).
a) What is the probability that the ball chosen from urn A is white, given that exactly 2 of the chosen balls are white?
b) Denote by X the chosen balls and throw them away. Next, choose a ball from each of the urns (uniformly at random) and denote those balls by Y. What *is the probability that Y contains exactly 2 white balls given that X contains exactly 1 white ball.
Sorry for the long post! I did section a) of the question and the final answer was 7/13. Not sure how to do the b) section. any answer or even a direction would be very helpful!
a)
Unfortunately I arrived at another outcome which made me decide to publish my calculation.
Let e.g. $(1,0,1)$ stand for the event that the balls chosen from urns A and C are white and the ball chosen from urn B is red. Then to be found is:$$P((1,-,-)\mid(1,1,0)\cup(1,0,1)\cup(0,1,1))=\frac{P((1,1,0)\cup(1,0,1))}{P((1,1,0)\cup(1,0,1)\cup(0,1,1))}$$The events are mutually exclusive so that:$$\dots=\frac{P((1,1,0))+P((1,0,1))}{P((1,1,0))+P((1,0,1))+P((0,1,1))}$$Applying independence we get: $$\cdots=\frac{\frac26\frac8{12}\frac34+\frac26\frac4{12}\frac14}{\frac26\frac8{12}\frac34+\frac26\frac4{12}\frac14+\frac46\frac8{12}\frac14}=\frac7{11}$$
b)
See my comment.