The probability of exactly two bonds defaulting over the next year is calculated using the Binomial distribution formula, which is applicable because the bond defaults are independent and identically distributed Bernoulli random variables. The formula for the probability of exactly k bond defaults (P(K=k)) is given by:

$P(K = k) = \frac{n!}{k!(n-k)!} \times p^k \times (1 - p)^{n-k}$

Where:

• $n$ is the total number of bonds (5 in this case),
• $k$ is the number of bond defaults we are interested in (2 in this case),
• $p$ is the probability of default of each individual bond (0.17 in this case),
• $(n-k)$ is the number of bonds that do not default.

Plugging in the values:

$P(K=2) = \frac{5!}{2!(5-2)!} \times 0.17^2 \times 0.83^3$

$P(K=2) = \frac{120}{2 \times 6} \times 0.0289 \times 0.5927$

$P(K=2) = 10 \times 0.0289 \times 0.5927$

$P(K=2) = 0.1652$

This calculation results in a probability of 16.52%, which corresponds to option C (16.5%), making it the correct answer.