1. The problem asks for the probability that exactly one new visitor arrives at a website in any given hour. 2. This is a classic Poisson distribution problem where the number of events (visitor arrivals) in a fixed interval follows the formula: $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda$ is the average number of arrivals per hour, $k$ is the exact number of arrivals we want the probability for, and $e$ is Euler's number (approximately 2.71828). 3. Since the problem does not specify $\lambda$, we assume it is known or given. For exactly one visitor, $k=1$. 4. Substitute $k=1$ into the formula: $$P(X=1) = \frac{\lambda^1 e^{-\lambda}}{1!} = \lambda e^{-\lambda}$$ 5. To find the numerical value, plug in the given or known $\lambda$ value. 6. Round the result to four decimal places as requested. Final answer: $$P(X=1) = \lambda e^{-\lambda}$$ (Note: Please provide the value of $\lambda$ to compute the exact probability.)