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Let a fisherman catch fish like a Poisson process s with known catching intensity λ=2 per 90 minutes.
The amount of fish caught at minute $t ∈ [0,90], s(t)$ is then Poisson distributed with expected value $E[s(t)]=\frac{tλ}{90},t ∈ [0,90]$.
The probability at time t to catch in 90 minutes exactly 1 fish, given that he caught no fish so far is $P(s(90)=1|s(t)=0)\space ,\space for\space t ∈ [0,90]$.Calculate the probability.

Anybody that can help me? Thanks in advance.

Cruceru Vlad
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    Welcome to Math.SE. This is a reasonably self-contained problem statement, but it lacks much context. The injunction to "Calculate the probability" will strike some Readers as indicating that you are passing along an assigned exercise without due effort to digest the problem statement yourself. We would prefer to see context such as what makes the problem important to you, or what difficulty you encountered in trying to solve it yourself. This will help Readers to respond in a fashion that is appropriate to your studies. – hardmath Sep 27 '18 at 14:11
  • Nicely put - might use that myself. – Paul Sep 27 '18 at 14:36

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If I understand correctly, the fisherman has caught $0$ fish after $t$ minutes (with $0\le t\le 90$). Given this information, we want to know what is the probability of at exactly one fish after a total of $90$ minutes has elapsed.

This means that we are looking for the probability of catching exactly one fish in $90-t$ minutes. As you note in your original post, the number of fish caught should follow a Poisson distribution with parameter proportional to the length of time of the interval.

Can you go from there?

paw88789
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