Answer: 0.084

## Explanation This is a Poisson distribution problem where: - λ (mean rate) = 3 accidents per day - We need P(N > 5) where N is the number of accidents **Step-by-step calculation:** 1. **Poisson probability formula:** \[ P(N=n) = \frac{e^{-\lambda} \lambda^n}{n!} \] 2. **Calculate P(N ≤ 5):** \[ P(N \leq 5) = P(N=0) + P(N=1) + P(N=2) + P(N=3) + P(N=4) + P(N=5) \] 3. **Individual probabilities:** - P(N=0) = e⁻³ × 3⁰/0! = e⁻³ × 1 = 0.0498 - P(N=1) = e⁻³ × 3¹/1! = 0.0498 × 3 = 0.1494 - P(N=2) = e⁻³ × 3²/2! = 0.0498 × 9/2 = 0.2240 - P(N=3) = e⁻³ × 3³/3! = 0.0498 × 27/6 = 0.2240 - P(N=4) = e⁻³ × 3⁴/4! = 0.0498 × 81/24 = 0.1680 - P(N=5) = e⁻³ × 3⁵/5! = 0.0498 × 243/120 = 0.1008 4. **Sum of probabilities for N ≤ 5:** \[ P(N \leq 5) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 = 0.9160 \] 5. **Calculate P(N > 5):** \[ P(N > 5) = 1 - P(N \leq 5) = 1 - 0.9160 = 0.0840 \] **Key concepts:** - Poisson distribution models rare events occurring at a constant average rate - λ = 3 represents the average number of accidents per day - The complement rule: P(N > 5) = 1 - P(N ≤ 5) - The answer 0.084 corresponds to option D

Author: Nikitesh Somanthe