statistics - Geometric distribution probability of no success at all

From S. Broverman, 2006:

A town's maintenance department has estimated that the cost of snow removal after a major

snowstorm is $100,000$. Historical information suggests that the number of major snowstorms in
a winter season follows a geometric distribution for which the probability of no major
snowstorms in a season is .4. The town purchases an insurance policy which pays nothing if
there is one or less major snowstorms in the season, but the insurance pays $50$% of all seasonal
snow removal costs for major snowstorms if there are $2$ or more major snowstorms. Find the
expected payout by the insurer.

I am confused by the bolded line "a geometric distribution for which the probability of no major
snowstorms in a season is $.4$". That sounds like it is saying the probability of no success at all is $.4$.

How does one find the probability of no success in a geometric distribution? From here it seems that one simply takes $(1-p)$ to the power of the $n$th attempt for which there is no success. However, according to that, how does one find the probability of no success at all, no matter how many attempts? Wouldn't that be $\lim \limits_{n \to \infty}(1-p)^n = 0$? If so, how would one understand the probability of no success being $.4$?

Answer

From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.

Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.

One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.

See also my question here for an analogous case.