Answer: 0.084

## 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**.

Author: Tanishq Prabhu