I will use the following notation to be as consistent as possible with the wiki (in case you want to go back and forth between my answer and the wiki definitions for the poisson and exponential.)
$N_t$: the number of arrivals during time period $t$
$X_t$: the time it takes for one additional arrival to arrive assuming that someone arrived at time $t$
By definition, the following conditions are equivalent:
$ (X_t > x) \equiv (N_t = N_{t+x})$
The event on the left captures the event that no one has arrived in the time interval $[t,t+x]$ which implies that our count of the number of arrivals at time $t+x$ is identical to the count at time $t$ which is the event on the right.
By the complement rule, we also have:
$P(X_t \le x) = 1 - P(X_t > x)$
Using the equivalence of the two events that we described above, we can re-write the above as:
$P(X_t \le x) = 1 - P(N_{t+x} - N_t = 0)$
But,
$P(N_{t+x} - N_t = 0) = P(N_x = 0)$
Using the poisson pmf the above where $\lambda$ is the average number of arrivals per time unit and $x$ a quantity of time units, simplifies to:
$P(N_{t+x} - N_t = 0) = \frac{(\lambda x)^0}{0!}e^{-\lambda x}$
i.e.
$P(N_{t+x} - N_t = 0) = e^{-\lambda x}$
Substituting in our original eqn, we have:
$P(X_t \le x) = 1 - e^{-\lambda x}$
The above is the cdf of a exponential pdf.
Hint: if you knew the average, you could determine the probability of having exactly 0, exactly 1, exactly 2, and so forth.
So, what you need to do is rearrange the equation that you use so that if you know what you know (probability of 1+ accidents =.98) you can determine the average.
Best Answer
What is the probability that at least two weeks will elapse between accident?
This uses the fact that time between Poisson events follows the exponential distribution (probability distribution that describes the time between events in a Poisson process - wikipedia./exponential distribution). When finding the probability of at least two weeks the lower bound is 2, and the distribution has a mean of 0.25. Hope this helps.