A four-digit identification card is made. Find the probability that the card will contain the digits $0,1,2,3$ in any order. (Round your answer to five decimal places.) The probability that the card will contain the digits $0,1,2,3$ in any order is 0.002

Step 1 ：The total number of four-digit numbers is \(10^4\) since each digit can be any number from 0 to 9.

Step 2 ：We are looking for the probability that the card will contain the digits 0,1,2,3 in any order.

Step 3 ：We need to calculate the number of ways these four digits can be arranged.

Step 4 ：This is a permutation problem, so we can use the formula for permutations of a multiset, which is \(n!\), where \(n\) is the number of items.

Step 5 ：In this case, \(n=4\), so there are \(4!=24\) ways to arrange the digits 0,1,2,3.

Step 6 ：Therefore, the probability is \(\frac{24}{10^4}\).

Step 7 ：Final Answer: The probability that the card will contain the digits 0,1,2,3 in any order is \(\boxed{0.0024}\).