# drawing balls from an urn (conditional probability)

Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?

I encountered this problem in my AoPS textbook, as a complete newcomer to conditional probability. Is the answer 1/2? I would really appreciate a solution, as I have no way to check my work.

-

Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn.
Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$
So the probability of drawing a red ball in the first draw give that knowledge is: $$P(D_1=r \mid D_2=k) \\ = P(D_1=r \cap U_1=a \mid D_2=k)+P(D_1=r \cap U_1=b \mid D_2=k) \\ = P(U_1=a)P(D_1=r \mid U_1=a \cap D_2=k)+P(U_1=b)P(D_1=r \mid U_1=b \cap D_2=k) \\ = \frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{3}{5} \\ = \frac{7}{15}$$