Explanation

This problem involves a Poisson distribution where the average rate of accidents is λ = 3 per day.

Poisson Probability Formula:

$P(N = n) = \frac{e^{-\lambda} \lambda^n}{n!}$

Step-by-step calculation:

We need to find P(N > 5), which equals: $P(N > 5) = 1 - P(N \leq 5)$ $P(N > 5) = 1 - [P(N = 0) + P(N = 1) + P(N = 2) + P(N = 3) + P(N = 4) + P(N = 5)]$

Individual probabilities:

• P(N = 0) = $\frac{e^{-3}3^0}{0!} = e^{-3} = 0.0498$
• P(N = 1) = $\frac{e^{-3}3^1}{1!} = 3e^{-3} = 0.1494$
• P(N = 2) = $\frac{e^{-3}3^2}{2!} = \frac{9e^{-3}}{2} = 0.2240$
• P(N = 3) = $\frac{e^{-3}3^3}{3!} = \frac{27e^{-3}}{6} = 0.2240$
• P(N = 4) = $\frac{e^{-3}3^4}{4!} = \frac{81e^{-3}}{24} = 0.1680$
• P(N = 5) = $\frac{e^{-3}3^5}{5!} = \frac{243e^{-3}}{120} = 0.1008$

Sum of probabilities: $P(N \leq 5) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 = 0.9160$

Final probability: $P(N > 5) = 1 - 0.9160 = 0.0840$

Therefore, the probability that more than 5 accidents occur on a single day is 0.084.