Expected value for Poisson distribution A busy intersection sees two traffic accident per month on average. Suppose the number of accidents follow a Poisson distribution. Starting from January next year, which month do you expect to be the first one in which there is no accident? 
I figured it would be the same for every month (as in every month is equally likely to have no traffic accidents), and can't figure out why a particular month would be expected as the first one to have no accidents. Any reason a single month is more likely than others? 

The probability that a month is accident-free is $e^{-2}$, and therefore the probability the month has at least one accident is $1-e^{-2}$.
Let $X$ be the number of the first month which is accident-free, where the months are numbered $1,2,3,\dots$, with January $1$ next year as Month $1$. Note that January $2$ years from now is Month $13$.
We have $\Pr(X=k)=(1-e^{-2})^{k-1} e^{-2}$.
Now we need to interpret the question. The random variable $X$ has geometric distribution, with $p$, the probability of "success" equal to $e^{-2}$.
The probability $\Pr(X=k)$ is a steadily decreasing function of $k$. So the most likely value of $\Pr(X=k)$ occurs at $k=1$. 
The expectation $E(X)$ of $X$, by the usual fact about the geometric distribution, is $\frac{1}{e^{-2}}$, that is, $e^2$. Since $e^2$ is not an integer, it is hard to assign a month to it. The value is about $7.4$. We could associate it with July. 
Because of the word "expect," it seems likely that we are expected to calculate $E(X)$. 
