The question involves the Poisson distribution, which is used to model the number of events (in this case, defaults) occurring in a fixed interval of time, given a constant average rate of occurrence. The average number of defaults per year is given as four.

The probability of a certain number of events occurring in a Poisson process can be calculated using the formula: $P(K = n) = \frac{e^{-\lambda} \lambda^n}{n!}$ where:

• $\lambda$ (lambda) is the average rate of occurrence (average number of defaults per year in this scenario),
• $n$ is the number of events we want to calculate the probability for,
• $e$ is the base of the natural logarithm.

The question asks for the probability of at most one default next year, which means we need to calculate the sum of the probabilities of zero defaults and one default occurring.

Using the Poisson formula:

• The probability of zero defaults ($P(K = 0)$) is: $P(K = 0) = \frac{e^{-4} \cdot 4^0}{0!} = \frac{e^{-4}}{1} = e^{-4} \approx 0.0183$
• The probability of one default ($P(K = 1)$) is: $P(K = 1) = \frac{e^{-4} \cdot 4^1}{1!} = \frac{4e^{-4}}{1} = 4e^{-4} \approx 0.0733$

Adding these probabilities together gives: $P(at most 1 default) = P(0 defaults) + P(1 default) = 0.0183 + 0.0733 = 0.0916$

However, the provided answer in the file content states 9.16%, which seems to be a rounding error. The correct calculation should yield approximately 9.16%, but the exact value from the calculation is 9.16% (0.0916 when converted from decimal to percentage). Therefore, the correct answer is 9.16%, which corresponds to option C.