Answer: Mean = 0.45, variance = 0.38

## 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.

Author: LeetQuiz .