Suppose that people come in according to a Poisson process of rate $\lambda = 20$ per hour. Suppose further that each person that arrives goes into house $A$ with probability $\frac{1}{3}$ and into house $B$ with probability $\frac{2}{3}$. Assume that a total of $100$ people arrived in the first $10$ hours. Conditional on this, I would like to find the probability that $n(n ≤ 100)$ people arrived into house A in the first $4$ hours.
Generally, I know that without the houses:
\begin{align} P(X(4)=N|X(10) = 100) &= \frac{P(X(4)=N,X(10) = 100)}{P(X(10) = 100)} \\ &= \frac{P(X(4)=N,X(10)-X(4) = 100-N)}{P(X(10) = 100)} \\ &= \frac{P(X(4)=N)\cdot P(X(10)-X(4) = 100-N)}{P(X(10) = 100)} \\ &= {100\choose N}\left(\frac{4}{6}\right)^N\left(\frac{6}{10}\right)^{100} \end{align}
To find the probability of $n$ people arriving into house $A$, I define that $N=N_1+N_2$, where $N$ is total people, $N_1$ people in house $A$, $N_2$ people in house $B$ and I have that:
\begin{align} P(X(4) = N, N_1=n|X(10)=100)&= \sum_{i=0}^{100-n}P(X(4) = N_1+N_2, N_1=n, N_2 = i|X(10)=100)\\ &= \sum_{i=0}^{100-n}P(X(4) = n+i, N_1=n, N_2 = i|X(10)=100)\\ \end{align}
At this point, I am not sure how to factor out the joint distribution of the last line. Am I on the right track?
Hint: The arrivals in house $A$ follow a Poisson process with parameter $λ_A=\frac1320$ per hour. Hence, you can ignore arrivals to house $B$ and only adapt your first part and your are done.
Note, that the last line of your first calculation in the first part seems to have two mistakes. It should be $$\dbinom{100}{N}\left(\frac4{10}\right)^N\left(\frac6{10}\right)^{100-N}$$