Well, yes, numbers can be repeated while being random. You need to do your own logic to validate if they're already on the array, this can be done with the following code:
In the code I used an array of 10 elements to observe there aren't repeated numbers even on that situation.
import java.util.Random;
public class RandomNumbersNoRepeating {
public static void main(String[] args) {
int array[] = new int[10];
Random random = new Random();
//Fills the array
for (int i = 0; i < array.length; i++) {
boolean found = false;
int r = 0;
do {
found = false;
r = random.nextInt(10) + 1;
//Here we check if the number is not on the array yet
for (int j = 0; j < array.length; j++) {
if (array[j] == r) {
found = true;
break;
}
}
} while (found);
array[i] = r;
}
//Prints the array
for (int i = 0; i < array.length; i++) {
System.out.println(array[i]);
}
}
}
Another possible solution, as given in a comment could be to shuffle an array from 1-10, and get the first four numbers