What's the probability that zero meteors were seen in the first hour but at least 10 were seen in the last three hours? (Poisson process)

Question

So the setup of the question is that a group of students is watching a meteor shower from $11$ p.m. to $3$ a.m. and they arrive according to a Poisson process with intensity $\lambda = 4$. The question is as stated above, "what's the probability that zero meteors were seen in the first hour but at least 10 were seen in the last three hours?" My thought was to add $$\begin{align*} P(N([0,1]) = 1) + P(N([1,4]) \geq 10) & = e^{-4}\left (\frac{0^{4}}{0!} \right ) + (1 - P(N([1,4]) <10) ). \\ & = e^{-4} + \left (1 - e^{-12}\sum_{k=0}^{9} \left (\frac{12^{k}}{k!} \right ) \right ). \\ & = 0.7759. \end{align*}$$ but the textbook says the correct answer is $0.01388$ so clearly I am misinterpreting the problem but I am not sure how else to interpret it. Any advice would be appreciated.

Answer 1

These are independent events. The probability of getting exactly $0$ in the first hour is: $Poisson[4,0]$.

The probability of seeing 0 to 9 (inclusive) is $\sum\limits_{i=0}^9 Poisson[12,i]$, where for the three-hour "unit" interval the mean rate is $12$. The probability of seeing $10$ or more is 1 minus that number.

Hence the total probability requested is:

$Poisson[4,0] \cdot \left( 1 - \sum\limits_{i=0}^9 Poisson[12,i] \right) = 0.0138762.$