1. **Stating the problem:** We want to find the probability that at least 5 accidents occur in a given month at an intersection. 2. **Understanding the problem:** "At least 5 accidents" means 5 or more accidents. If we denote the number of accidents by the random variable $X$, we want $P(X \geq 5)$. 3. **Formula used:** If $X$ follows a Poisson distribution with mean $\lambda$, then $$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ for $k = 0, 1, 2, \ldots$. 4. **Important rule:** The probability of at least 5 accidents is $$P(X \geq 5) = 1 - P(X < 5) = 1 - P(X \leq 4) = 1 - \sum_{k=0}^4 P(X = k)$$ 5. **Intermediate work:** Calculate $$\sum_{k=0}^4 P(X = k) = \sum_{k=0}^4 \frac{e^{-\lambda} \lambda^k}{k!}$$ 6. **Final step:** Substitute the value of $\lambda$ (mean number of accidents per month) if given, compute the sum, then subtract from 1. 7. **Example:** If $\lambda = 3$, then $$P(X \geq 5) = 1 - \sum_{k=0}^4 \frac{e^{-3} 3^k}{k!}$$ Calculate each term: $$P(X=0) = e^{-3} \frac{3^0}{0!} = e^{-3}$$ $$P(X=1) = e^{-3} \frac{3^1}{1!} = 3e^{-3}$$ $$P(X=2) = e^{-3} \frac{3^2}{2!} = \frac{9}{2} e^{-3}$$ $$P(X=3) = e^{-3} \frac{3^3}{3!} = \frac{27}{6} e^{-3} = \frac{9}{2} e^{-3}$$ $$P(X=4) = e^{-3} \frac{3^4}{4!} = \frac{81}{24} e^{-3} = \frac{27}{8} e^{-3}$$ Sum these and subtract from 1: $$P(X \geq 5) = 1 - e^{-3} \left(1 + 3 + \frac{9}{2} + \frac{9}{2} + \frac{27}{8}\right)$$ Simplify inside the parentheses: $$1 + 3 = 4$$ $$\frac{9}{2} + \frac{9}{2} = 9$$ $$4 + 9 = 13$$ $$13 + \frac{27}{8} = \frac{104}{8} + \frac{27}{8} = \frac{131}{8}$$ So, $$P(X \geq 5) = 1 - e^{-3} \times \frac{131}{8}$$ Calculate numerically: $$e^{-3} \approx 0.0498$$ $$P(X \geq 5) \approx 1 - 0.0498 \times 16.375 = 1 - 0.8157 = 0.1843$$ **Answer:** The probability that at least 5 accidents occur in a month is approximately $0.1843$ (rounded to four decimal places).