Ok here's the question:
Fisherman Dan is out fishing by a stream. On average, 3 fishes per hour swim by but Fisherman Dan catches the fish with probability 1/2. It rains in average once per day. The fish and the rain arrive according to independent Poisson processes.
Given that fisherman Dan caught exactly 2 fishes between 8 and 12 o'clock, what is the probability that he caught zero fish during the first hour of fishing?
This was my initial solution:
\begin{align*} & Pr(\text{2 fish caught during 8-12 o'clock | no fish caught during 8-9 o'clock}) = \\ = &Pr(N(12)\; - \;N(8)\; = 2\;|\; N(9)\;-\;N(8) = 0 \;) =\\ = &Pr(N(4)\; = 2\;|\; N(1) = 0 \;) = \\ = & Pr(N(4)\; = 2\;)\cdot Pr(\; N(1) = 0 \;) = \\ = & e^{-6}\dfrac{6^2}{2!} \cdot e^{-3/2} = \\ = &0.00995 \end{align*}
But in the answer sheet they've reasoned in this way instead (translated so keep that in mind):
If the only thing we know is the number of events under a certain time intervall and these events stem from a Poisson process, then we also know that these events are independent and identically distributed over the time intervall. Given two events occurred during 4 hours we obtain:
Pr(no event during the first hour) = Pr(both events during the last three hours) = $(\dfrac{3}{4-0})^2 = \dfrac{9}{16} = 0.5625$
why doesn't my method work though? I understand their solution, but mine should work, right?
You computed P(2 fish 8-12 | 0 fish 8-9) and you needed P(0 fish 8-9 | 2 fish 8-12)