Here's the problem:
The number of calls per hour arriving at an answering service follows a Poisson process with $\lambda = 4$. It is known that exactly "k" calls arrived in the first four hours. What is the probability that exactly j of them arrived in the first hour?
I think I'm stumped on the setup. I know its a conditional probability but the events aren't independent, correct? Also, I know how to add Poisson rvs, and that x4 can be changed to be a poisson with $\lambda = 16$. $$ P(x_1 = j | x_4 = k) ? $$
While it is true that the arrival count in the first hour is not independent of arrival count in the whole four hours, we can say that the arrival count in the first hour is independent of the count in the last three hours. These periods are non-overlapping, and Poisson events are memoryless.
If exactly $j$ arrive in the first hour, $k-j$ need to arrive in the next three in order for $k$ to arrive in the given four hour period.
$\begin{align} \text{Let } X_4 & = X_1 + X_3 & {X_1: \text{arrivals in the first hour}\\X_3: \text{arrivals in the last three hours } \\ X_4: \text{arrivals in all four hours}} \\[2ex] \therefore \mathsf P(X_1 = j\mid X_4=k) & = \frac{\mathsf P(X_1=j \cap X_3=k-j)}{\mathsf P(X_4=k)} & {\text{due to the partitioning}} \\[1ex] & = \frac{\mathsf P(X_1=j)\mathsf P(X_3=k-j)}{\mathsf P(X_4=k)} & \text{because of independence} \\[1ex] & = \frac{f(j; 4)\,f(k-j; 12)}{f(k; 16)} & {\text{where } f(x; \lambda) = \dfrac{\lambda^x e^{-\lambda}}{x!}} \\[1ex] &\ddots \end{align}$
When you perform the necessary substitutions and simplify, what do you notice about the resulting expression? It should look rather familiar.