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.
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?