Is this a trick question? Balls into urns probability There are two urns. Urn 1 contains 3 white and 2 red ball, urn 2 one white and two
red. First, a ball from urn 1 is randomly chosen and placed into urn 2. Finally, a ball from
urn 2 is picked. This ball be red: What is the probability that the ball transferred from
urn 1 to urn 2 was white?
My answer is 3/5 - 3W over 5 balls in urn 1 as the second event does not tell me anything to influence which ball was transferred since urn 2 already has a red ball anyways. Hope to know if my reasoning is valid!
 A: Method I: (Bayes) There are two scenarios in which a red ball is observed:
I:  A white ball is transferred (probability $\frac 35$).  In this case, a red ball is observed with probability $\frac 12$.  Thus this scenario has probability $$\frac 35 \times \frac 12 = \frac 3{10}$$
II:  A red ball is transferred (probability $\frac 25$).  Now the probability of drawing a red ball is $\frac 34$  Thus this scenario has probability $$\frac 25\times \frac 34 =\frac 3{10}$$
We see that the two scenarios contribute equally, thus the probability that it was a white ball that was transferred initially is $\boxed {\frac 12}$
Note:  as our prior was that the probability the transferred was white was $\frac 35$ we see that the observation of the red ball has caused us to lower our estimate for the probability
Method II (Conditional Probability).  A priori, the universe here consists of four possible events:  $(W,W),(W,R),(R,R),(R,W)$ according to the color of the transferred ball and the color of the drawn ball.  A routine calculation shows that  $(W,W),(W,R),(R,R)$ each have probability $\frac 3{10}$ and $(R,W)$ has probability $\frac 1{10}$.  We are asked for the probability that the transferred ball is $W$ conditioned on the fact that the drawn ball is $R$ and inspection now shows the answer to be $\frac 12$.  
A: Confirming lulu's answer.
You begin with ($3$white and $2$red) balls in urn#1, then select $1$ at random(without bias) and move it into urn#2.   Urn#2 then contains either ($2$white and $2$red) or ($1$white and $3$red) depending on which colour of ball was moved into it (white or red respectively).   Then a ball is extracted(without bias) from urn#2 and it turns out to be red.
Let $W_1$ be the event that a white ball was moved into urn#2, and $R_2$ the event that a red ball was extracted from urn#2.   We wish to determine $\mathsf P(W_1\mid R_2)$, the conditional probability that a white ball had been transferred into urn#2 when given that a red ball was subsequently extracted.
We know $~\mathsf P(W_1)=3/5~,~ \mathsf P(R_2\mid W_1)=2/4~,~ \\\mathsf P(W_1^\complement)=2/5~,~ \mathsf P(R_2\mid W_1^\complement)=3/4~$.
Then we apply Bayes' Rule:$$\begin{align}\mathsf P(W_1\mid R_2) ~&=~\dfrac{\mathsf P(W_1)\,\mathsf P(R_2\mid W_1)}{~\mathsf P(W_1)\,\mathsf P(R_2\mid W_1)+\mathsf P(W_1^\complement)\,\mathsf P(R_2\mid W_1^\complement)~}\\[1ex]&=~\dfrac{3\cdot 2}{~3\cdot 2+2\cdot 3~}\\[1ex]&=~\dfrac{~1~}{2}\end{align}$$
$\blacksquare$
A: 
A Pretty Straightforward question. Please go through the diagram. Another way to think is if 1W transferred to U2= (1W + 2R) makes U2 = (2R + 2W). Now, # White Balls = # Red Balls . Then P(W)= P(R) = 1/2 for Urn2.
