# Conditional probability involving Poisson random variable

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 $\lambda = 16$. $$P(x_1 = j | x_4 = k) ?$$

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}
• @ittiekat That's right. ${k\choose j}\frac{4^j\,12^{k-j}}{16^k} = {k\choose j}\left(\frac{1}{4}\right)^j\left(\frac{3}{4}\right)^{k-j}$. This is the probability of $j$ from $k$ arrivals in the period being in its first quarter. Commented Nov 7, 2014 at 20:29