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