Probability, balls from boxes Consider two boxes: there are 15 white and 12 black balls in the first box, and 14 white
and 18 black balls in the second box. Anna puts
her hand in the first box, takes at once two balls and places them in the second box. Then,
she takes one ball wihout looking from the second box.

Knowing that she took a black ball from the second box, what is the probability that
  she transferred two balls of different colors from the first box to the second box?

I tried using Bayes Theorem:
X:=transfer 2 colours
Y:= pick black from the 2nd box
$P(X|Y)=\frac{P(Y|X)(P(X)}{P(Y)}$
with
$P(X)=\frac{(15,2)(12,0)}{(27,2)}+\frac{(15,0)(12,2)}{(27,2)}$
$P(Y)=\frac{18}{32}$
$P(Y|X)=\frac{18}{34}+\frac{20}{34}$
Does this work like this?
 A: You can use simple conditional probability i.e.:
$$P(X|Y)=\frac{P(X\cap Y)}{P(Y)}$$
Where in this case it would be:
$$P(X\cap Y) = P(\text{transfer diff colours})*P(\text{picking black|transfered diff colours})$$
$$P(X\cap Y) =  \left( 2. \frac{15}{27}\frac{12}{26}\right)\frac{19}{34} = \frac{190}{663}$$
for $P(Y)$ we can use total probability:
$$P(Y) =  \left( \frac{12}{27}\frac{11}{26}\right) \frac{20}{34}+ \left( \frac{15}{27}\frac{14}{26}\right)\frac{18}{34}  +  \left(2. \frac{15}{27}\frac{12}{26}\right)\frac{19}{34} = \frac{5}{9}$$
Result is therefore: $P(X|Y)=\frac{114}{221}\approx0.5158..$
A: To summarize callculus' suggestion:
X1: 1b(black), 1w(white) from the 1st box. X2: 2b from 1st. X3: 2w from 1st.
Y: 1b from the 2nd box.
Then P(X1|Y)=$\frac{P(Y|X1)P(X1)}{P(Y)}$, where:
P(Y|X1)= $\frac{19}{34}$
P(Y) = $\sum_{i=1}^3 P(Y|Xi)P(Xi)$
P(X1)= $\frac{C_1^{15}C_1^{12}}{C_2^{27}}$
P(X2)= $\frac{C_2^{15}}{C_2^{27}}$
P(X3)= $\frac{C_2^{12}}{C_2^{27}}$
Hope I didn't type something wrong. You should be able to figure out the computations.
BTW, when you find any probability greater than 1, something is definitely wrong.
A: HINT  
I would say
$P(X) = \frac{20}{39}$ (see colleagues 'calculus' comment)
$P(Y|X)=\frac{19}{34}$
Knowing that she took a black ball from the second box: $\Rightarrow P(Y)=1$
$ \Rightarrow P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)}=\frac{P(Y|X)P(X)}{1}=\cdots$
