Explanation

This is a Poisson distribution problem. The key information:

• Portfolio has 10 independent bonds
• Average of 1 default in 5 years
• We need probability of exactly 1 default in 1 year

Step 1: Calculate the annual default rate (λ)

• Average defaults per year = 1 default / 5 years = 0.2 defaults per year
• So λ = 0.2

Step 2: Apply Poisson probability formula The Poisson probability formula for exactly k events is:

$P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}$

For exactly 1 default (k = 1): $P(X = 1) = \frac{e^{-0.2} \cdot 0.2^1}{1!}$ $P(X = 1) = e^{-0.2} \cdot 0.2$

Step 3: Calculate the probability

• e^{-0.2} ≈ 0.81873
• 0.81873 × 0.2 = 0.163746
• Convert to percentage: 0.163746 × 100 = 16.37%

Therefore, the probability of exactly one default in a year is 16.37%, which corresponds to option A.

Note: The fact that there are 10 bonds is not needed for this calculation since we're given the average default rate directly. The Poisson distribution assumes defaults are independent and rare events, which fits this scenario.