If I have a bitarray of size $m$ with all $0$'s initially and one universal hash function $h: U \rightarrow [m]$ where each index in the bitarray has $\frac{1}{m}$ probability of being flipped to 1, then the probability of any index $i$ in two bitarrays $x$ and $y$ having the same bit flipped is: $P[h(x_i)=h(y_i)]=\frac{1}{m}$.
Now, if I have a binary tree like shown below where the root is just the union of all inserted bit arrays, the right subtree is the union of b1 and b2, and the left tree is the union of b1, then the probability of choosing the right subtree for insertion (based on the bitwise AND) will be greater. For instance, if I were to insert b4=00100 into the below tree then the probability of choosing right subtree will be $\frac{2}{5}$ (although the bitwise AND is $0$ for both subtrees in this case).
How can I find the expected number of insertions before a subtree get's full? I know that the expected number of insertions before one bit array gets full is $\sum_{i}^{m}\mathbb{E}(X_i) =\sum_{i}^{m}\frac{1}{m} = m$, but the tricky part for me is that there is a probability of either subtrees getting picked and this would increase the expectation (I think).
insertions: b1=10000, b2=01000, b3=00001 based on bitwise AND
11001 \\root with union of all bit arrays. b1 is left, while b2 and b3 is right
/ \
/ \
10000 01001