1. **Problem statement:** We have ID numbers starting with letter S, followed by 2 digits representing a year, then 6 digits. A. How many different ID cards can be made in a year? B. How many cards can be made in a year if the last 3 digits are all 0's? C. What is the probability that a randomly selected student has the last 3 digits of their ID as 0's? 2. **Formula and rules:** - Each digit can be from 0 to 9, so 10 possibilities per digit. - The letter S is fixed. - For counting total IDs, multiply possibilities for each digit. - Probability = (favorable outcomes) / (total outcomes). 3. **Step A: Total different ID cards in a year** - The year digits: 2 digits, each 10 possibilities, so $10 \times 10 = 100$. - The last 6 digits: each 10 possibilities, so $10^6 = 1,000,000$. - Total IDs in a year = $1 \times 100 \times 1,000,000 = 100,000,000$. 4. **Step B: Cards with last 3 digits all 0's** - Last 3 digits fixed as 0, so only 1 possibility. - The first 3 digits after S (2 year digits + first 3 of last 6) have $10^3 = 1,000$ possibilities. - Total such IDs = $1 \times 100 \times 1,000 = 100,000$. 5. **Step C: Probability last 3 digits are 0's** - Probability = (number with last 3 digits 0) / (total number) - $$\text{Probability} = \frac{100,000}{100,000,000} = \frac{1}{1,000} = 0.001$$ **Final answers:** A. $100,000,000$ B. $100,000$ C. $0.001$ (or 0.1%)