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.