1. **Problem Statement:** We have ID numbers starting with letter S, followed by 2 digits for the year, then 6 digits. A. How many different ID cards can be made in a year? B. How many cards can be made if the last 3 digits are all zeros? C. What is the probability that a randomly selected student has last 3 digits as zeros? 2. **Formulas and Rules:** - Each digit can be from 0 to 9, so 10 possibilities per digit. - For counting total IDs, multiply possibilities for each digit. - Probability = (Number of favorable outcomes) / (Total number of outcomes). 3. **Solution:** **A. Number of different ID cards in a year:** - The year digits are fixed for a given year, so only the last 6 digits vary. - Each of the 6 digits can be 0-9, so total possibilities = $10^6$. **B. Number of cards with last 3 digits all zeros:** - Last 3 digits fixed as 0, so only first 3 digits of the 6 vary. - Each of these 3 digits can be 0-9, so possibilities = $10^3$. **C. Probability that last 3 digits are zeros:** - Probability = (Number of IDs with last 3 digits zero) / (Total IDs) - $$\text{Probability} = \frac{10^3}{10^6}$$ - Simplify by canceling common factors: $$\frac{\cancel{10^3} \times 1}{\cancel{10^3} \times 10^3} = \frac{1}{10^3}$$ - So, probability = $\frac{1}{1000}$. 4. **Final answers:** - A: $1000000$ - B: $1000$ - C: $\frac{1}{1000}$