I figured out a rather complicated program that computes the answer, but I'm not sure how to explain it. If it's not clear from my question, my grasp on stats and math notation is tenuous at best. (Sorry.)
The hard part of the problem comes down to how to represent the intersection of multiple trials. Once that part is down, we can use $$P(E_n|\bigcap_{j=1}^{n-1}E_j)=P(\bigcap_{j=1}^{n}E_j)/P(\bigcap_{j=1}^{n-1}E_j)$$.
I couldn't find a single generalizable way to do it, but I was able to narrow it down to two cases: the simple case and the... less simple case. The example in the question demonstrates the simple case.
Each trial can be written as a set of tuples that encode the number of red balls in $D_j$ and $D_j^C$. Let $T$ be a set of tuples in the form $(B, r)$ where $B$ is a subset of the balls in the urn, and $r$ is the number of red balls in $B$. So trial 1 is $T_1 = \{(\{b_1,b_2\}, 1),(\{b_3,b_4,b_5\},2)\}$ and trial 2 is $T_2 = \{(\{b_1,b_2,b_3\}, 2),(\{b_4,b_5\},1)\}$.
I'm not sure how this works, but I was able to find $T_n \cap T_m$ by taking $T_n \cup T_m$ and transforming it using the following process:
- Compare each element with each other element, looking for elements where the first tuple value is a proper subset of the first tuple value of another element. In other words, find a $t = (B_a, r_a)$ where there exists some $s = (B_b, r_b)$ where $t, s \in T$ and $B_b \subset B_a$.
- Replace $t$ in the set with $(B_a - B_b, r_a - r_b)$.
- Repeat steps (1) and (2) until for every $t, s \in T$ where $t = (B_a, r_a)$ and $s = (B_b, r_b)$, $B_b \not\subset B_a$.
In our example, $T_1 \cup T_2 = \{(\{b_1,b_2\}, 1),(\{b_3,b_4,b_5\},2),(\{b_1,b_2,b_3\}, 2),(\{b_4,b_5\},1)\}$. We see an element $t = (\{b_1,b_2,b_3\}, 2)$ and another element $s = (\{b_1,b_2\}, 1)$. We can replace $t$ with $(\{b_3\}, 1)$. We continue until we end up with $T_1 \cap T_2 = \{(\{b_1,b_2\}, 1),(\{b_3\},1),(\{b_4,b_5\},1)\}$.
Then check if there are any elements whose first value intersects with the first value of any other element. If for every $(B_a, r_a), (B_b, r_b) \in T$ $B_a$ and $B_b$ are disjoint sets, then we are in the simple case. Otherwise, we are in the less simple case.
Simple case: all sets disjoint
Because all the sets in the tuples in the simple case are disjoint, each $t \in T$ can be considered a draw from the urn without replacement. To find $P(T)$, consider a $D$ that is a sequence of draws $d_i$ from the urn without replacement where each $d_i = (B_i,r_i)$ is a different element of T. (Basically, pretend $T$ is ordered.)
$$P(T) = \prod\limits_{i=1}^{|T|}P(X_i=r_i)$$
where $X_i \sim HypG(m_i,N_i,n_i)$
$n_i = |B_i|$
$m_i = m_{i-1} - r_{i - 1}$
$N_i = N_{i - 1} - n_{i - 1}$
$N_1 =$ number of balls in the urn
$m_1 =$ number of red balls in the urn.
In other words, take the probability that $d_i$ has $r_i$ red balls, update the total number of balls and red balls left to account for the draw, and repeat until the urn is empty.
Recall that for a hypergeometric distribution, $P(X_i=r_i)=\frac{\dbinom{m_i}{r_i}\dbinom{N_i - m_i}{n_i - r_i}}{\dbinom{N_i}{n_i}}$, and you have a solution for the simple case. Going back to the example $P(T_1 \cap T_2) = \left(\frac{\dbinom{3}{1}\dbinom{2}{1}}{\dbinom{5}{2}}\right)\left(\frac{\dbinom{2}{1}\dbinom{1}{0}}{\dbinom{3}{1}}\right)\left(\frac{\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{2}{2}}\right) = 2/5$. Since $P(T_1) = 3/5$, $P(T_2|T_1) = 2/3$.
Less simple case: some sets intersecting
But imagine a less simple case where for some $(B_a, r_a), (B_b, r_b) \in T$, $B_a$ intersects with $B_b$. Then $P(T) = \sum\limits_{b \in B_a}P(T|E_b)P(E_b)$ where $E_b$ is the event that $b$ is red. In other words, we add up the probabilities $T$ happens given $b$ is red times the probability that $b$ is red. (Law of total probability.) We know to encode $E_b$ as $T_b = \{(\{b\},1)\}$ and repeat the whole song and dance for $T \cap T_b$.
For example, consider a
$T_1 = \{(\{b_1,b_2\}, 1),(\{b_3,b_4,b_5\},2)\}$ and
$T_2 = \{(\{b_1,b_3\},1),(\{b_2,b_4,b_5\},2)\}$.
We find that $T_1 \cap T_2 = \{(\{b_1,b_2\}, 1),(\{b_1,b_3\},1),(\{b_2,b_4,b_5\},2),(\{b_3,b_4,b_5\},2)\}$.
Since $\{b_1,b_2\}$ intersects with $\{b_1,b_3\}$, we take $\{b_1,b_2\}$ and create two cases: $R_1 = \{(\{b_1\}, 1)\}$ and $R_2 = \{(\{b_2\}, 1)\}$. We then see that $P(T_1 \cap T_2) = P(T_1 \cap T_2|R_1)P(R_1) + P(T_1 \cap T_2|R_2)P(R_2)$.
To represent $T_1 \cap T_2|R_i$, remove 1 red ball at $b_i$ from the urn. So you end up with
$T_1 \cap T_2|R_1 = \{(\{b_2\}, 0),(\{b_3\},0),(\{b_4,b_5\},2)\}$
$T_1 \cap T_2|R_2 = \{(\{b_1\}, 0),(\{b_3\},1),(\{b_4,b_5\},1)\}$
$P(T_1 \cap T_2 | R_1) = \left(\frac{\dbinom{2}{0}\dbinom{2}{1}}{\dbinom{4}{1}}\right)\left(\frac{\dbinom{2}{0}\dbinom{1}{1}}{\dbinom{3}{1}}\right)\left(\frac{\dbinom{2}{2}\dbinom{0}{0}}{\dbinom{2}{2}}\right) = 1/6$
$P(T_1 \cap T_2 | R_2) = \left(\frac{\dbinom{2}{0}\dbinom{2}{1}}{\dbinom{4}{1}}\right)\left(\frac{\dbinom{2}{1}\dbinom{1}{0}}{\dbinom{3}{1}}\right)\left(\frac{\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{2}{2}}\right) = 1/3$
Note that $P(R_1) = P(R_2) = 3/5$, so $P(T_1 \cap T_2) = (3/5)(1/6 + 1/3) = 3/10$ and $P(T_2|T_1) = 1/2$.