Originally Posted by

**crazy_gal108**

**3.** Six students are asked to secretly choose a number from 1 to 15. Determine the probability that at least two students choose the same number to the nearest thousandth.

Since we are talking 'bout thousands only the first three digits are important. Thus, all the numbers are,

Code:

1.000
1.001
1.002
........
14.998
14.999
15.000

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 $\displaystyle 14000/14001$

STUDENT3)Can chose any number except that of student1 and student2 thus the probability is $\displaystyle 13999/14001$

STUDENT4)Probability is $\displaystyle 13998/14001$

STUDENT5)Probability is $\displaystyle 13997/14001$

STUDENT6)Probability is $\displaystyle 13996/14001$

Thus, the probability that they DO is,

$\displaystyle 1-\frac{_{14000}\mbox{P}_{5}}{14001^5}$

Evaluating we find that,

0.0010709185167321222566663866769298

Is the probability thus,

a little more than 1%