Since the defaults are independent events with a constant probability, the number of defaults follows a binomial distribution. Using the binomial probability formula: $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$

Where: $n = 3$ (number of bonds) $k = 1$ (number of defaulting bonds) $p = 0.10$ (probability of default) $1-p = 0.90$ (probability of non-default)

$P(X = 1) = \binom{3}{1} \cdot (0.10)^1 \cdot (0.90)^2 = 3 \cdot 0.10 \cdot 0.81 = 0.243$.