# Conditional Probability

• Oct 12th 2009, 08:25 PM
abbasb
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!
• Oct 12th 2009, 08:36 PM
matheagle
You need to compute $P(A|B)={P(AB)\over P(B)}$

Where
A=ball chosen from urn A was white
and
B=exactly 2 white balls were selected.
• Oct 12th 2009, 08:48 PM
abbasb
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)
• Oct 12th 2009, 09:49 PM
matheagle
I don't know about all that A-complement 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... $W_1R_2W_3$ OR $W_1W_2R_3$

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 ... $R_1W_2W_3$
• Oct 13th 2009, 05:37 PM
Math Quest