The placement of the non-red balls is irrelevant...the problem is decided by the placement of the red ones alone.
To do the first part, it is easier to work from the complimentary event. The probability that a specified red ball goes to a bin other than the second is $\frac 34$. Thus the answer is $$1-\left( \frac 34\right)^2=\boxed {\frac 7{16}}$$
To do the second one, it might be helpful to simply list the possible placements of the two red balls. Given that the second bin must contain a red ball, there are only four cases: $$(1,1,0,0)\quad (0,2,0,0)\quad (0,1,1,0)\quad (0,1,0,1)$$
Where writing, say, $(1,1,0,0)$ means that the first two bins each have one red ball and the second two each have none.
Note that the probability of getting $(0,2,0,0)$ is $\frac 1{16}$ since we must have both red balls going to the second bin. The probability of getting, say, $(1,1,0,0)$ is $\frac 18$ since we can get this configuration in two ways (either the first red ball goes in the first bin and the second red ball goes in the second bin, or the first red ball goes in the second bin and the second red ball goes in the first bin). This gives another way to see $\frac 7{16}$ as the result for the first problem, since $$\frac 18+\frac 1{16}+\frac 18+\frac 18=\boxed {\frac 7{16}}$$
We see from this that the conditional probability that the first bin has a red ball (conditioned on the fact that the second bin also has one) is $$\frac {1/8}{7/16}=\boxed {\frac 27}$$
Side note: if you prefer to work with equi-probable scenarios (not a bad idea) then you need to indicate the placement of $r_1, r_2$ (the two red balls) separately. Thus the scenario $(1,1,0,0)$, say, becomes two scenarios, $(r_1,r_2,0,0)$ and $(r_2,r_1,0,0)$. You wind up with seven scenarios in which the second bin contains at least one red ball, each of which having probability $\frac 1{16}$.
In my opinion, the events should be
- $X$: The number of white balls drawn from Urn 2; the three possible outcomes 2,1, and 0 define the partition you need.
- $B$: The ball drawn from Urn 1 is white; this is the conditioning event.
The probabilities we are after are then $P(B)$ and $P(X=1|B)+ P(X=2|B)$.
Regarding the former, we use the law of total probability to say that
$$P(B) = P(B|X=2)P(X=2) + P(B|X=1) P(X=1) +P(B|X=0)P(X=0).$$
Where
$$P(X=2) = 1/4$$
$$P(X=1) = 1/2$$
$$P(X=0) = 1/4$$
and where
$$P(B|X=n) = \frac{1+n}{4}$$
(If I have transferred $n$ white balls, I have $n+1$ white balls in a total of $4$.)
Once you have these, you can use Bayes' theorem to find $P(X=n|B)$ and the result:
$$P(X=1\vee X=2|B) = 1-P(X=0|B) = 1- \frac{P(B|X=0)P(X=0) }{P(B)};$$
where we have used the fact that $X=n$ is a partition: the probabilities of the events $X=1\vee X=2$ and $X=0$ sum to one
$$P(X=1\vee X=2|B) + P(X=0|B) =1.$$
Note that you don't need to observe this and you can directly compute:
$$P(X=1\vee X=2|B) = \frac{P(B|X=1)P(X=1) + P(B|X=2)P(X=2) }{P(B)};$$
where we have used the fact that $P(X=1\wedge X=2) =0$. The result will be the same.
Best Answer
You can combine like that because there are ten balls in each bin and an equal choice between bins, therefore each ball has an equal chance of being selected. Nine of the twenty balls are red.
If there were different amounts in each bin, there would be bias, and you would instead have to use the Law of Total Probability to get the correct result. However, that is not the case, so you are okay.
Of course, you may still use it to obtain the same answer:
$$\begin{align}\mathsf P(r)&=\mathsf P(r\mid B_1)\mathsf P(B_1)+\mathsf P(r\mid B_2)\mathsf P(B_2)\\[1ex]&= \frac 2{10}\cdot\frac 12+\frac 7{10}\cdot\frac 12\\[1ex]&=\frac 9{20}\end{align}$$
So, yes, indeed. The probability for having selected bin 1 when given that you obtained a red ball is $2/9$. It is less likely that you picked from the bin with fewer red balls -- two among the nine red balls come from bin 1 and all balls are each equally likely to have been selected.
$$\begin{align}\mathsf P(B_1\mid r)&=\dfrac{\mathsf P(r\mid B_1)\mathsf P(B_1)}{\mathsf P(r\mid B_1)\mathsf P(B_1)+\mathsf P(r\mid B_2)\mathsf P(B_2)}\\[1ex]&=\frac 29\end{align}$$