What is the Probability of Divisibility by 3?

[Ackbach]

Here is this week's POTW:

-----

Three numbers from the first 99 positive integers are chosen at random (with repetitions allowed). What is the probability that the sum is divisible by 3?

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to kiwi for his correct solution, which follows:

Spoiler

When we choose a number between 1 and 99 that number will be congruent to 0,1 or 2 mod 3. There are 33 numbers in each congruence class. So there is equal probability of selecting 0,1 or 2 mod 3.

When we select three such numbers there are 3^3=27 possible outcomes mod 3. I write abc to represent the first selection being in class a, the second selection being in class b and the third outcome being in class c.

The sum of the three outcomes is 0 mod 3 if one of the following outcomes occurs abc=:
000
012
021
102
201
120
210
222
111

The sum of the three outcomes is 1 mod 3 if one of the following outcomes occurs:
001
010
100
022
202
220
112
121
211

The sum of the three outcomes is 2 mod 3 if one of the following outcomes occurs:
002
020
200
011
101
110
221
212
122

This exhausts the 27 possibilities and 9/27 outcomes have a sum congruent to 0 mod 3.

There is a 1/3 probability of the sum of the numbers being congruent to 0 mod 3 so there is a 1/3 probability of the sum of the selections being divisible by 3.