A security alarm requires a four-digit code. The code can use the digits \( 0-9 \) and the digits cannot be repeated. Which expression can be used to determine the probability of the alarm code beginning with a number greater than 7 ? \( \frac{\left({ }_{2} \mathrm{P}_{1}\right)\left({ }_{9} \mathrm{P}_{3}\right)}{{ }_{10} \mathrm{P}_{4}} \) \( \frac{\left({ }_{2} C_{1}\right)\left({ }_{9} C_{3}\right)}{{ }_{10} C_{4}} \) \( \frac{\left({ }_{10} \mathrm{P}_{1}\right)\left({ }_{9} \mathrm{P}_{3}\right)}{{ }_{10} \mathrm{P}_{4}} \) \( \frac{\left({ }_{10} C_{1}\right)\left({ }_{9} C_{3}\right)}{{ }_{10} C_{4}} \)

To solve this probability problem, consider that the alarm code must start with a digit greater than 7—in this case, that means either 8 or 9. Once the first digit is chosen, you have 9 remaining digits (00-09, excluding the first choice) to fill the next three positions without repeating digits. This leads to the correct probability expression as follows: The expression \( \frac{\left({ }_{2} \mathrm{P}_{1}\right)\left({ }_{9} \mathrm{P}_{3}\right)}{{ }_{10} \mathrm{P}_{4}} \) accurately reflects this because it calculates the number of successful outcomes (choosing from 2 possible starting digits and arranging the remaining 3 from the 9) against the total number of possible arrangements (4 digits from 10 choices). Want to get a bit deeper into probabilities? This type of problem involves combinatorial mathematics, which dates back centuries. Think of mathematicians like Blaise Pascal who laid the groundwork for modern probability. Learning about their lives and discoveries can be an adventure in itself! If you're fascinated by how these concepts apply beyond the classroom, think about how they affect real-world statistics and data analysis! They help determine things like security system vulnerabilities or even lottery odds. Understanding these principles can empower you to make more informed decisions in various aspects of life—like whether to trust that flashy new security system!