a security alarm requires a four - digit code. the code can use the digits 0 - 9 and the digits cannot be…

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?\n$\frac{(_{2}p_{1})(_{9}p_{3})}{_{10}p_{4}}$\n$\frac{(_{2}c_{1})(_{9}c_{3})}{_{10}c_{4}}$\n$\frac{(_{10}p_{1})(_{9}p_{3})}{_{10}p_{4}}$\n$\frac{(_{10}c_{1})(_{9}c_{3})}{_{10}c_{4}}$\n

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?\n$\frac{(_{2}p_{1})(_{9}p_{3})}{_{10}p_{4}}$\n$\frac{(_{2}c_{1})(_{9}c_{3})}{_{10}c_{4}}$\n$\frac{(_{10}p_{1})(_{9}p_{3})}{_{10}p_{4}}$\n$\frac{(_{10}c_{1})(_{9}c_{3})}{_{10}c_{4}}$\n

Answer

Explanation:

Step1: Calculate 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)!}), where (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 number of four - digit codes starting with a number greater than 7

Numbers greater than 7 in the set ({0,1,\cdots,9}) are 8 and 9. So there are 2 choices for the first digit. After choosing the first digit, we need to choose the remaining 3 digits from the remaining 9 digits. The number of ways to choose the remaining 3 digits is ({9}P{3}=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7). The number of four - digit codes starting with a number greater than 7 is (({2}P{1})\times({9}P{3})) (using the multiplication principle, where ({2}P{1}=\frac{2!}{(2 - 1)!}=2)).

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}})