Since we are talking 'bout thousands only the first three digits are important. Thus, all the numbers are,
Originally Posted by crazy_gal108
To make the problem easier, let us do the opposite statement. Meaning, the probability that 6 students do not choose the same number and then from that subtract one.
There are a total of 14,001 numbers.
STUDENT1)Can chose any one thus probability is 1.
STUDENT2)Can chose any number except that of student1, thus the probability is
STUDENT3)Can chose any number except that of student1 and student2 thus the probability is
Thus, the probability that they DO is,
Evaluating we find that,
Is the probability thus,
a little more than 1%