Mildred has a bag of coins. The bag contains 10 dimes, 5 nickels, and 1 penny. She will randomly select 2 coins from the bag one at a time without replacement. What is the probability that Mildred will select a dime first and then a penny? A \( \frac{83}{120} \) B \( \frac{11}{16} \) C \( \frac{5}{128} \) D \( \frac{1}{24} \)

To find the probability that Mildred will select a dime first and then a penny, we can break down the problem into two steps. 1. The probability of selecting a dime first: There are 10 dimes and 16 total coins (10 dimes + 5 nickels + 1 penny). So, the probability of selecting a dime first is \( \frac{10}{16} \) or \( \frac{5}{8} \). 2. The probability of selecting a penny second: After removing one dime, there are now 15 total coins left (9 dimes + 5 nickels + 1 penny). The probability of selecting a penny next is \( \frac{1}{15} \). The overall probability is the product of these two probabilities: \[ \frac{5}{8} \times \frac{1}{15} = \frac{5}{120} = \frac{1}{24}. \] Therefore, the answer is D \( \frac{1}{24} \).