# Poisson distribution huge comfusion

I have this problem... although I tried a few failed attempts but..

The number of boats that being repaired at a repair campus , during a week, is a Poisson distribution with k parameter . If k is also a random variable that gets the values 1,2,3,4 with probability 0.3 , 0.2 , 0.4 and 0.1 respectively, what is the probability that no ship is repaired this week?

So, I don't understand how to solve this, since we are not given the rate (k). Also I am a bit confused, because k is also a variable. I mean I don't exactly know how to attack this question.

I tried to solve it supposing that k variable is x ,in order to find k but its was helpless try .

Any opinions regarding this issue ?

Thank you

-

No boats repaired can happen in $4$ (disjoint) ways: (i) $k=1$ and no boats are being repaired or (ii) $k=2$ and no boats are being repaired or (iii) $k=3$ and no boats are being repaired or (iv) $k=4$ and $\dots$.
The probability that a Poisson with parameter $\lambda$ is $0$ is equal to $e^{-\lambda}$. So the probability of (i) is $(0.3)e^{-1}$.
Similarly, the probability of (ii) is $(0.2)e^{-2}$.