Given that the number of hurricanes per year follows a Poisson distribution with a mean $\lambda = 2.35$, we can calculate the probability of having no hurricanes ($X = 0$) using the formula for the Poisson distribution:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$

For $k = 0$, the formula simplifies to:

$P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$

Now, we substitute $\lambda = 2.35$ into the equation:

$P(X = 0) = e^{-2.35}$

Let me calculate this.The probability of having no hurricanes next year is approximately $0.0953$.

Thus, the correct answer is:

$\boxed{0.0953}$

Do you have any questions or want further details?

Here are 5 related questions to expand your understanding:

1. What is the probability of exactly one hurricane next year?
2. How would the probability change if the mean number of hurricanes increased to 3.5?
3. What is the cumulative probability of having 2 or fewer hurricanes next year?
4. How does the Poisson distribution apply to rare events in real-world scenarios?
5. How can you interpret the parameter $\lambda$ in the context of a Poisson distribution?

Tip: In a Poisson distribution, the mean ($\lambda$) and variance are equal, making it useful for modeling events that happen at a constant rate.