Explanation

This is a binomial distribution problem where:

• Number of trials (n) = 3 bonds
• Probability of success (default) for each trial (p) = 0.15
• The events are independent

Mean Calculation: For a binomial distribution, the mean (expected value) is: $\mu = n \times p = 3 \times 0.15 = 0.45$

Variance Calculation: For a binomial distribution, the variance is: $\sigma^2 = n \times p \times (1-p) = 3 \times 0.15 \times (1-0.15) = 3 \times 0.15 \times 0.85 = 0.3825$

Rounding to two decimal places, variance = 0.38

Verification:

• Option A: Mean = 0.15 (incorrect), Variance = 0.32 (incorrect)
• Option B: Mean = 0.45 (correct), Variance = 0.38 (correct) ✓
• Option C: Mean = 0.45 (correct), Variance = 0.32 (incorrect)
• Option D: Mean = 0.15 (incorrect), Variance = 0.38 (incorrect)

Therefore, the correct answer is B with mean = 0.45 and variance = 0.38.