In urn I there are $3$ white balls and $2$ black balls. Urn II contains $2$ white balls and $3$ black balls.
- We draw a ball from urn I and either throw it back or put it into urn II - each of these two possibilities has probability $\frac{1}{2}$.
- We then do the same for urn II.
What is the probability that the first ball drawn was white, if it is known that after two draws urn II contains exactly two white balls?
I made a tree as shown here:
From that I think it's correct to say that the possibility is equal to: $$\begin{align*} &\frac{\frac{2}{5} \cdot \frac{1}{2} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2} \cdot \frac{2}{6} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2} \cdot \frac{4}{6} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2} \cdot \frac{4}{6} \cdot \frac{1}{2}}{\frac{3}{5} \cdot \frac{1}{2} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{3}{5} \cdot \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{1}{2} + \frac{3}{5} \cdot \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{1}{2} + numerator} \\[0.3cm] = &\frac{2 \cdot \frac{2}{5} + 2 \cdot \frac{3}{5} + 2 \cdot \frac{3}{5} + 2 \cdot \frac{2}{6} + 2 \cdot \frac{4}{6} + 2 \cdot \frac{4}{6}}{3 \cdot \frac{2}{5} + 3 \cdot \frac{3}{5} + 3 \cdot \frac{3}{5} + 2 \cdot \frac{2}{5} + 2 \cdot \frac{3}{5} + 2 \cdot \frac{3}{5} + 2 \cdot \frac{2}{6} + 2 \cdot \frac{4}{6} + 2 \cdot \frac{4}{6}} \\[0.3cm] = &\frac{\frac{4}{5} + \frac{6}{5} + \frac{6}{5} + \frac{4}{6} + \frac{8}{6} + \frac{8}{6}}{\frac{6}{5} + \frac{9}{5} + \frac{9}{5} + \frac{4}{5} + \frac{6}{5} + \frac{6}{5} + \frac{4}{6} + \frac{8}{6} + \frac{8}{6}} \\[0.3cm] = &\frac{\frac{16}{5} + \frac{10}{3}}{\frac{40}{5} + \frac{10}{3}} = \frac{48 + 50}{120 + 50} = \frac{49}{85} \end{align*}$$
Is there any easier / more elegant way to calculate that?
Your answer ($49/85$) appears to incorrect. Here's one way we could approach this problem.
Let $X$ be the event of the first ball drawn being white. Also, let $Y$ be the event of there being exactly $2$ white balls in urn II at the end.
We need to find $P(X|Y)$. Using Bayes' theorem and the law of total probabilities, we have
$$P(X|Y) = \frac{P(X) \cdot P(Y|X)}{P(X) \cdot P(Y|X) + P(X^c) \cdot P(Y|X^c)}$$
Now, $P(X) = 3/5$ and $P(X^c) = 2/5$. This was the easy part.
1. Calculating $P(Y|X)$
We break this into two cases,
Case a. The first ball is put back into urn I
If the first ball drawn is white and that ball is put back into urn I, there is only one way in which the urn II will NOT end up with exactly $2$ white balls - the second ball drawn is white and it is put into urn I. The probability of that happening is $ \dfrac{2}{5} \cdot \dfrac{1}{2} = \dfrac{1}{5}$.
So, the total probability for this case would be
$$\frac{1}{2} \cdot \left[ 1 -\frac{1}{5} \right] = \frac{2}{5}$$
Multiplied by $1/2$ because that's the probability of the first ball being put back in urn I.
Case b. The first ball is put into urn II
If the first ball drawn is white and that ball put into urn II, there is only one way in which the urn II will end up with exactly $2$ white balls - the second ball drawn is white and it is put into urn I. The probability of that happening is $ \dfrac{3}{6} \cdot \dfrac{1}{2} = \dfrac{1}{4}$.
So, the total probability for this case would be
$$\frac{1}{2} \cdot \frac{1}{4}= \frac{1}{8}$$
Combining the two cases, we have
$$P(Y|X) = \frac{2}{5} + \frac{1}{8} = \frac{21}{40}$$
2. Calculating $P(Y|X^c)$
If the first ball drawn is black and that ball put into urn II, there is only one way in which the urn II will NOT end up with exactly $2$ white balls - the second ball drawn is white and it is put into urn I.
Again, we break this into two cases,
Case a. The first ball is put back into urn I
The probability of that second ball drawn is white and it is put into urn I is $ \dfrac{2}{5} \cdot \dfrac{1}{2} = \dfrac{1}{5}$.
So, the total probability for this case would be
$$\frac{1}{2} \cdot \left[ 1 -\frac{1}{5} \right] = \frac{2}{5}$$
Case b. The first ball is put into urn II
The probability of that second ball drawn is white and it is put into urn I is $ \dfrac{2}{6} \cdot \dfrac{1}{2} = \dfrac{1}{6}$.
So, the total probability for this case would be
$$\frac{1}{2} \cdot \left[ 1 -\frac{1}{6} \right]= \frac{5}{12}$$
Combining the two cases, we have
$$P(Y|X^c) = \frac{2}{5} + \frac{5}{12} = \frac{49}{60}$$
Now that we have everything we need, substituting the values into the formula, we get
$$\begin{align*} P(X | Y) &= \frac{\frac{3}{5} \cdot \frac{21}{40}}{\frac{3}{5} \cdot \frac{21}{40} + \frac{2}{5} \cdot \frac{49}{60}} \\[0.3cm] &=\frac{189}{385} \approx 0.49 \end{align*}$$