Answer: 9.16%

## Explanation Using the Poisson distribution with λ = 4 defaults per year, we calculate the probability of at most one default (0 or 1 default): **Probability of 0 defaults:** \[ P(K = 0) = \frac{(4)^0}{0!} \cdot e^{-4} = 1 \cdot e^{-4} = 0.0183 \] **Probability of 1 default:** \[ P(K = 1) = \frac{(4)^1}{1!} \cdot e^{-4} = 4 \cdot e^{-4} = 0.0733 \] **Total probability (at most one default):** \[ P(K \leq 1) = P(K = 0) + P(K = 1) = 0.0183 + 0.0733 = 0.0916 = 9.16\% \] This demonstrates the application of Poisson distribution to model rare events like defaults, where λ represents the average rate of occurrence.

Author: LeetQuiz .