Urn 1 contains N balls (with N even and large), half of them are white and half of them are black. Those balls are transferred to Urn 2, but some of them fall out in the process. In particular, a white ball succesfully move from Urn 1 to Urn 2 with probability 0.8, while a black ball succesfully move with probability 0.3. How do I calculate the probability that, if I pick a random ball from Urn 2, this ball is white? Note that I do not know how many balls ended up in Urn 2.
EDIT What is this probability in the limit for $N \rightarrow \infty$?
Would the calculation be different if, instead of saying that half of the balls in Urn 1 are whithe and half are black, I say that each ball in Urn 1 has 0.5 probability of being black and 0.5 probability of being white?
Look at the finite case with $N = 2n$ balls, of which there are $n$ balls of each color in Urn 1. The number of white balls $W$ successfully transferred to the second urn is a binomial random variable: $$W \sim \operatorname{Binomial}(n, p_1 = 0.8).$$ Similarly, the number of black balls $B$ successfully transferred to the second urn is binomial: $$B \sim \operatorname{Binomial}(n, p_2 = 0.3).$$
Therefore, the unconditional probability that a randomly selected ball drawn from the second urn is white, is given by $$\begin{align} \pi_n = \sum_{w = 0}^n \sum_{b = 0}^n \frac{w}{w+b} \Pr[W = w]\Pr[B = b] &= \sum_{w=0}^n \sum_{b=0}^n \frac{w}{w+b} \binom{n}{w} \binom{n}{b} p_1^w (1-p_w)^{n-w} p_2^b (1-p_2)^{n-b}. \end{align}$$ This sum does not appear to have an elementary closed form. It is a function of $n$, $p_1$, and $p_2$. For the specified probability parameters, we can compute a table: $$\begin{array}{c|c} n & \pi_n \\ \hline 1 & 0.68 \\ 2 & 0.7552 \\ 3 & 0.755768 \\ 4 & 0.750153 \\ 5 & 0.745753 \\ 6 & 0.742663 \\ 7 & 0.740436 \\ 8 & 0.738767 \\ 9 & 0.737472 \\ 10 & 0.736439 \\ 11 & 0.735595 \\ 12 & 0.734893 \\ 13 & 0.7343 \\ 14 & 0.733792 \\ 15 & 0.733353 \\ 16 & 0.73297 \\ 17 & 0.732631 \\ 18 & 0.732331 \\ 19 & 0.732062 \\ 20 & 0.731821 \\ \end{array} $$ and a plot for larger $n$ shows:
It is not difficult to see that in the limit as $n \to \infty$, the probability is simply $$\frac{p_1}{p_1 + p_2} = \frac{8}{11}.$$
If the distribution of colors in the first urn is not deterministic but is itself a binomial random variable, i.e. with $N$ balls, of which $\mathcal W_0 \sim \operatorname{Binomial}(N, p_0 = 0.5)$ are white, then there is yet another layer of conditioning involved. The probability becomes $$\pi_N = \sum_{k = 0}^N \sum_{w = 0}^k \sum_{b=0}^{N-k} \frac{w}{w+b} \binom{k}{w}\binom{N-k}{b} p_1^w (1-p_1)^{k-w} p_w^b (1-p_w)^{N-k-b} \binom{N}{k} 2^{-N}.$$