Answer-first summary for fast verification
Answer: 9.16%
C is correct. Using the Poisson distribution approach, and assuming the average number of defaults is \( A \) per year, the probability of \( n \) defaults over a period (year) \( t \) is given as: \[ P(K = n) = \frac{e^{-\lambda} \lambda^n}{n!} \] Therefore, \( P(\text{at most one default}) = P(\text{one default}) + P(\text{no default}) \) \[ P(\text{at most one default}) = \left( \frac{4 \times 1}{1!} \right) e^{-4} + \left( \frac{4 \times 1}{0!} \right) e^{-4} = 0.0733 + 0.0183 = 9.16\% \]
Author: LeetQuiz Editorial Team
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A fixed-income analyst is examining the annual default rate within a bond portfolio and has determined that it follows a Poisson process. Historical data suggests that, on average, there are four bond defaults per year. Considering that these defaults occur independently of one another, calculate the probability that there will be at most one default in the next year.
A
6.58%
B
7.33%
C
9.16%
D
25.00%
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