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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?

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    $\begingroup$ The probability of the wrong event was computed incorrectly. The first solution multiplies the probability of $2$ fish in $4$ hours by the probability of no fish in the first hour. If the events were independent, which they are not, ths would give us the probability of $2$ fish in $4$ hours and $0$ fish in one hour. But the problem asks us to compute a conditional probability, which is something else entirely. $\endgroup$ – André Nicolas May 20 '15 at 15:18
  • $\begingroup$ yeah I could see why the two events are not independent. Thus it doesn't satisfy the increments ---> not a poisson counting process? $\endgroup$ – DOGOFWALLSTREET May 20 '15 at 15:22
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    $\begingroup$ I would suggest going back to fundamentals, and writing out a solution as follows. Let $A$ be the event no fish in first hour, and let $B$ be the event $2$ fish in $4$ hours. We want $\Pr(A| B)$, which is $\Pr(A\cap B)/\Pr(B)$. The calcuation of $\Pr(B)$ is easy. For $\Pr(A\cap B)$, this is the probability of no fish in first hour times the probability of $2$ fish in the next three hours. $\endgroup$ – André Nicolas May 20 '15 at 15:48
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You computed P(2 fish 8-12 | 0 fish 8-9) and you needed P(0 fish 8-9 | 2 fish 8-12)

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  • $\begingroup$ well, by stationary and independent increments, it would be the same calculation. $\endgroup$ – DOGOFWALLSTREET May 20 '15 at 15:15
  • $\begingroup$ $P(A|B) \neq P(B|A)$, even here $\endgroup$ – gt6989b May 20 '15 at 15:16
  • $\begingroup$ fair enough, but calculating P(0 fish 8-9 | 2 fish 8-12) instead, I would still use the increments to reach a solution. So what do I need to do instead? $\endgroup$ – DOGOFWALLSTREET May 20 '15 at 15:19
  • $\begingroup$ and I conditioned the way I did because I figured why would you condition on the end-result, to find out the probability of what could have happened during a period in the time interval? $\endgroup$ – DOGOFWALLSTREET May 20 '15 at 15:27
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    $\begingroup$ @DOGOFWALLSTREET because knowing what happened over a large interval influences information you know about its subintervals. $\endgroup$ – gt6989b May 20 '15 at 15:39

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