Poisson Distribution Fishes swim by a fisherman at a rate of approximately 3 fishes per hour. The fisherman has 1/2 chance of catching each fish. The fishes arrival rate independent and poisson distributed. The question is, what the chance that in 4 hours of fishing, our fisherman manages to catch atleast 2 fishes? My method of solving it is this: P(atleast 2 fishes) = 1 - P(fisherman gets 0 fishes) - P(fisherman get 1 fish). The answer that is given to us in the solution is: $$1 - e^{-6} -6e^{-6}$$ What i am wondering and want an answer from you guys is the following: What is the equation for P(fisherman gets 0 fishes) and P(fisherman gets 1 fishes) and how would it look if we wanted to find for for example 3 fishes or "n" fishes?

You have a Poisson distribution $p(n)$, where $n$ is the number of fish caught in an interval, with a mean per interval (here the rate of fish, i.e. "fish caught per interval, on average") $\mu=\frac{3}{2} \text{ per hour}\times 4 \text{ hours}=6$ when the interval is $4$ hours. Then the probability of catching $n$ fish in the interval is simply $$p(n)=\frac{\mu^n e^{-\mu}}{n!}$$