Question

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 { ( _ { 2 } p _ { 1 } ) ( _ { 9 } p _ { 3 } ) } { _ { 1 0 } p _ { 4 } } \\)

\\( \frac { ( _ { 2 } c _ { 1 } ) ( _ { 9 } c _ { 3 } ) } { _ { 1 0 } c _ { 4 } } \\)

\\( \frac { ( _ { 1 0 } p _ { 1 } ) ( _ { 9 } p _ { 3 } ) } { _ { 1 0 } p _ { 4 } } \\)

\\( \frac { ( _ { 1 0 } c _ { 1 } ) ( _ { 9 } c _ { 3 } ) } { _ { 1 0 } c _ { 4 } } \\)

Explanation:

Step1: Calculate the total number of four - digit codes

The total number of four - digit codes using digits \(0 - 9\) without repetition is given by the permutation formula \(_{n}P_{r}=\frac{n!}{(n - r)!}\). Here \(n = 10\) and \(r=4\), so the total number of four - digit codes is \(_{10}P_{4}=\frac{10!}{(10 - 4)!}=\frac{10!}{6!}=10\times9\times8\times7\).

Step2: Calculate the number of four - digit codes starting with a number greater than 7

Digits greater than 7 are 8 and 9. So, the number of ways to choose the first digit (a number greater than 7) is \(_{2}P_{1}=\frac{2!}{(2 - 1)!}=2\) (using the permutation formula since order matters for the first digit).
After choosing the first digit, we have 9 remaining digits to choose the remaining 3 digits from. The number of ways to choose and arrange the remaining 3 digits is \(_{9}P_{3}=\frac{9!}{(9 - 3)!}=9\times8\times7\).
The number of four - digit codes starting with a number greater than 7 is \((_{2}P_{1})(_{9}P_{3})\)

Step3: Calculate the probability

The probability \(P\) of an event is the number of favorable outcomes divided by the number of total outcomes. So the probability of the alarm code beginning with a number greater than 7 is \(\frac{(_{2}P_{1})(_{9}P_{3})}{_{10}P_{4}}\)

Answer:

\(\frac{(_{2}P_{1})(_{9}P_{3})}{_{10}P_{4}}\) (the first option)