
Conditional Probability
Hey
I've just been stuck on this question for a while and can't seem to figure out how to do it. I would appreciate any help.
Problem:
Consider 3 urns. Urn A contains 2 white and 4 red balls, Urn B contains 8 white and 4 red balls, and Urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected.
Thanks!

You need to compute $\displaystyle P(AB)={P(AB)\over P(B)}$
Where
A=ball chosen from urn A was white
and
B=exactly 2 white balls were selected.

Sorry, but I don't know how to format my symbols to get them to look like yours but,
I did do that and I got
P(AlB) = P(BlA)(PlA)/(P(BlA)P(A) + P(BlAc)P(Ac))
Where Ac is complement of A.
And I dont really know what to put for P(BlA) and P(BlAc)

I don't know about all that Acomplement stuff
JUST write down what B and AB are....
and figure out the probability of those events
FOR example, the event AB isn't so hard.
It consists of two mutually exclusive events...$\displaystyle W_1R_2W_3$ OR $\displaystyle W_1W_2R_3$
where the subscript denotes which urn.
BY independence and since we are only selecting ONE marble, that probability is
(2/6)(4/12)(1/4)+(2/6)(8/12)(3/4)
B consists of those two events and only one more ...$\displaystyle R_1W_2W_3$

Answer for your question