# Set numbers

• Jul 21st 2012, 01:36 PM
donnagirl
Set numbers
A number a is to be chosen at random from the set of {1,2,3,4,5,6}. A number b is to be chosen at random from the remaining five numbers. What is the probability that a/b will be an integer?

How can I prove that the answer is indeed 4/15?
• Jul 21st 2012, 01:55 PM
Plato
Re: Set numbers
Quote:

Originally Posted by donnagirl
A number a is to be chosen at random from the set of {1,2,3,4,5,6}. A number b is to be chosen at random from the remaining five numbers. What is the probability that a/b will be an integer?
How can I prove that the answer is indeed 4/15?

Well there thirty pairs $(a,b)$ from the cross product such that $a\ne b$.
You can list the pairs $(a,b)$ such that $\frac{a}{b}\in \mathbb{N}.$ Then count them.
• Jul 21st 2012, 02:30 PM
donnagirl
Re: Set numbers
Is there more of a proof way or combinatorial way to approach then just brute force counting?
• Jul 21st 2012, 02:53 PM
Plato
Re: Set numbers
Quote:

Originally Posted by donnagirl
Is there more of a proof way or combinatorial way to approach then just brute force counting?

I don't what to say that. With such a small outcome what would anyone want to do it the hard way?
Let $d_n$ be the number of divisors of $n$ less than $n$.
$d_1=0,~d_2=1,~d_3=1,~d_4=2,~d_5=1,~d_6=3$
There are eight such pairs out of thirty pairs.

If your set were $\{1,2,3,\cdots,98,99\}$ then the answer is: $\frac{{\sum\limits_{n = 2}^{99} {{d_n}} }}{{(99)(98)}}$
• Jul 21st 2012, 03:02 PM
emakarov
Re: Set numbers
I am not sure if this simplifies the counting, but you can select b first and then a. This does not change the answer. For b = 1 there are 5 multiples of b among {1, 2, 3, 4, 5, 6} - {b}, for b = 2 there are 2 multiples, and for b = 3 there is 1 multiple. For other b's there are no multiples distinct from b in this set. The number of "successful" pairs (a, b) is 5 + 2 + 1 = 8.
• Jul 22nd 2012, 12:11 AM
richard1234
Re: Set numbers
You may have to just list the ordered pairs (a,b) based on the value of b:

(2,1) (3,1) (4,1) (5,1) (6,1)
(4,2) (6,2)
(6,3)

There are 6P2 = 30 ways to select a,b and 8 of them work. The probability is 8/30 = 4/15.
• Jul 22nd 2012, 06:36 AM
Wilmer
Re: Set numbers
Code:

a:  1  2  3  4  5    6 b:  0  1  1  1,2  1  1,2,3  : 8     5  5  5  5  5    5    :30