Question

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.

Explanation:

Step1: Calculate total number of four - digit combinations

Each digit of the four - digit identification card can be from \(0\) to \(9\). So the total number of four - digit combinations is \(n = 10\times10\times10\times10=10^{4}\)

Step2: Calculate number of favorable combinations

We want to use the digits \(0\), \(1\), \(2\), \(3\). The number of permutations of \(n\) distinct objects taken \(r\) at a time is \(P(n,r)=\frac{n!}{(n - r)!}\). Here \(n = 4\) (the four digits \(0\), \(1\), \(2\), \(3\)) and \(r=4\). But we need to consider that a four - digit number cannot start with \(0\).
The total number of permutations of the four digits \(0\), \(1\), \(2\), \(3\) is \(A_{4}^4=\frac{4!}{(4 - 4)!}=4! = 24\). The number of permutations that start with \(0\) is \(A_{3}^3=\frac{3!}{(3 - 3)!}=3! = 6\). So the number of favorable four - digit numbers (using \(0\), \(1\), \(2\), \(3\)) is \(m=4!-3!=24 - 6=18\)

Step3: Calculate the probability

The probability \(P=\frac{m}{n}\). Substitute \(m = 18\) and \(n = 10000\) into the formula. \(P=\frac{18}{10000}=0.00180\)

Answer:

\(0.00180\)