Poisson Distribution, not sure how to include several events This is the very last question on an assignment sheet and for some reason I can't wrap my head around the very last sentence without doubting my approach. Here's the question:

Tony’s home has a number of electric lights, which fail from time to
  time and have to be replaced. On average Tony has to replace 1.5 light
  bulbs every month.
(i) What is the appropriate distribution of the total number of light
  bulbs which need to be replaced each year? What are its mean and
  variance?
(ii) What is the probability that Tony will have to replace at least
  one light bulb in each of the next three months?

For (i) I've said that I think it's Poisson Distribution and both the mean and variance will be 18 for a full year.
For (ii) I've written down the following:
Let $ \lambda$ be $1.5$, $P(X \ge 1) = 1 - P(X = 0)$
$$ P(X = 0) = {e^{1.5} \cdot 1.5^0 \over 0!}$$
$$= 1 - 0.3347\ldots$$
I'm not even sure if that's on the right track, but it occured to me that I have to somehow account for the last part:

in each of the next three months.


We are invited to assume that the lifetimes of bulbs have exponential distribution, which is a somewhat unreasonable assumption. But never mind, we will hold our noses and think Poisson.
The number of failures in a month has Poisson distribution, mean $1.5$.  
We have $3$ independent Poisson random variables, $X_1$, $X_2$, and $X_3$, which give us the number of failures in the first month, the second, the third.
We want the probability that $X_1\gt 0$ and $X_2\gt 0$ and $X_3\gt 0$.
Note that $\Pr(X_i\gt 0)=1-e^{-1.5}$.  
