Probabilities/Poisson distribution exercise I'm new to this site and I'm having trouble while solving a probability problem.
The question is: We have a regional airport, and the number of planes that arrive on the airport follow a poisson distribution of y=2
What is the probability of at least 3 planes takeoff in at least 3 of 5 days?

I was thinking about hipergeometrical probability, but I really have no idea how to solve this. I'd apreciate a lot if someone could give me a hand on this.

EDIT: The answer is 0,195
We have

Poisson$(\lambda,k)$

$$p(X=k)={\lambda^k\over k!}e^{-\lambda}.$$

The Poisson part tells us via the measurement for one unit of time, that $y=2$.The average of a poisson is $1\over\lambda$, so it's not clear to me whether $\lambda=2$ or ${1\over2}$. Is 2 the average number of planes, or is 2 the decay rate of the distribution? Let's assume there's 2 planes a day so $\lambda={1\over2}$. 

Above, it looks like you found $P(k\ge3\text{ planes in one day})$. (It's easier to calculate the complement?)

Let's notate that with a $p$ for success. Then P(at least 3 days)=P(3)+P(4)+P(5) as you say. And 

$$P(k)={5\choose k}p^k(1-p)^{5-k}.$$
We have

Poisson$(\lambda,k)$

$$p(X=k)={\lambda^k\over k!}e^{-\lambda}.$$

The Poisson part tells us via the measurement for one unit of time, that $y=2$.The average of a poisson is $1\over\lambda$, so it's not clear to me whether $\lambda=2$ or ${1\over2}$. Is 2 the average number of planes, or is 2 the decay rate of the distribution? Let's assume there's 2 planes a day so $\lambda={1\over2}$. 

Above, it looks like you found $P(k\ge3\text{ planes in one day})$. (It's easier to calculate the complement?)

Let's notate that with a $p$ for success. Then P(at least 3 days)=P(3)+P(4)+P(5) as you say. And 

$$P(k)={5\choose k}p^k(1-p)^{5-k}.$$