The problem involves two selections one after the other. In the first selection we pick out a set of objects. And then in the second selection we pick out a particular object from this set.
Consider we have 3 sacks each containing $n_1$ red balls (distinct shades), $n_2$ blue balls (distinct shades) and $n_3$ green balls (distinct shades) respectively. Consider $3$ independent random variables $X_1, X_2, X_3$ associated with the number of balls picked from each sack respectively (withput replacement) i.e $X_1$ is the number of red balls picked ($0 \leq X_1 \leq n_1)$. The probabilities of picking a ball associated with each colours are $p_1,p_2,p_3$ respectively.
Now that we have defined how to pick a set of balls which itself is random, we define how to pick a ball of particular color and particular distinct shade form this set or not pick a ball at all.
The probability of picking a ball of particular distinct shade and colour, say a particular shade of RED without loss of generality (present in the set is): $$q_{rl} =\frac{1}{\epsilon + X_1+X_2+X_3}$$
and not picking any ball is:
$$q_n=\frac{\epsilon}{\epsilon + X_1+X_2+X_3}$$
Now what we want to find is the probability of picking up a ball which is RED (shade doesn't matter). Thus the probability for that will be: $$\sum_l q_{rl}$$ i.e the sum of all probabilities of picking any shade of red denoted by $l$ from the constructed set.
What we need is the value of $\mathbb E \big[\sum_l q_{rl} \big]$ i.e the expectation of the probability that the final ball in our hand is red.
It is given that:
$$\mathbb E \big[\sum_l q_{rl}\big] = \mathbb E\bigg[\frac{D_1}{\epsilon + D_1+D_2+D_3}\bigg]$$
where:
$$D_i = Binomial(n_i,p_i)$$
Intuitively this seems clear but I cannot show this formally. I also could not find any resorce to deal with division of random variables except this, which seems insufficient here.
This is intuitive because, one way to write the probability of picking a ball of particular distinct shade and colour from our random set is (say RED):
$$q_{rl} =\frac{C_{rl}}{\epsilon + \sum_{i=1}^{n_1}C_{ri} + \sum_{j=1}^{n_2}C_{bj} + \sum_{k=1}^{n_3}C_{rk}}$$
where $C_{rl}$ will be $Ber(p_1)$, similarly if we had $C_{bl}$ it would be $Ber(p_2)$. Thus what we have is that picking a ball of particular color (say RED) is sum of Bernoulli random variables divided by sum of Bernoulli random variables $\frac{X_1}{\epsilon + X_1+X_2+X_3}$ which intuitively should be related to Binomial random variables, somehow.
