Let $X \sim \text{Poisson}(\lambda)$ be a random variable which counts the number of events that occur in a certain time interval. Suppose that each event can be classified as either type $A$ or type $B$, and that these events occur, independently of all other events, with probability $p$ and $1-p$, respectively. Let $X^{(A)}$ and $X^{(B)}$ denote the number of events of type $A$ and $B$, respectively. Show that $X^{(A)}$ and $X^{(B)}$ are independent random variables, and find their marginal distributions.
My progress:
So far I have been able to show that $X^{(A)} \sim \text{Poisson}(\lambda p)$ and $X^{(B)} \sim \text{Poisson}((1-p)\lambda)$. Now I'm trying to show that $X^{(A)}$ and $X^{(B)}$ are independent. By definition of independence, we want to show that
$$P(X^{(A)} = x_A, X^{(B)} = x_B) = P(X^{(A)} = x_A) P(X^{(B)} = x_B).$$
The right-hand side is easily computed since we have found the marginal distributions of $X^{(A)}$ and $X^{(B)}$. We have \begin{align*} P(X^{(A)} = x_A) P(X^{(B)} = x_B) &= \frac{(p \lambda)^{x_A} e^{-p \lambda}}{x_A!} \frac{((1-p)\lambda)^{x_B} e^{-(1-p)\lambda} }{x_B!} \\[5pt] &= \frac{\lambda^{x_A + x_B} p^{x_A} (1-p)^{x_B} e^{{-}\lambda} }{x_A! x_B!}. \end{align*}
But now I'm stuck. How do I compute $P(X^{(A)} = x_A, X^{(B)} = x_B)$?
Begin with the fact that $X=X^{(A)}+X^{(B)}$. $$\begin{align}\mathsf P(X^{(A)}=a, X^{(B)}=b) & =\mathsf P(X^{(A)}=a, X=a+b)\\[1ex]&=\mathsf P(X=a+b)\,\mathsf P(X^{(A)}=a\mid X=a+b)\\[1ex]&=\dfrac{\lambda^{a+b}\,\mathrm e^{-\lambda}}{(a+b)!}\cdot \mathsf P(X^{(A)}=a\mid X=a+b)\end{align}$$
Now argue that the conditional distribution for the count of type-A arrivals, when given an amount of arrivals, will be $\underline{\phantom{\textrm{binomial}}}$. Since the event that a particular arrival is of type-A is independent of the events that the other arrivals in the sequence are of type-A.$$X^{(A)}\mid X~\sim~\phantom{\mathcal {Bin}(X,p)}$$