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Math Help - Set numbers

  1. #1
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    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?
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  2. #2
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    Re: Set numbers

    Quote Originally Posted by donnagirl View Post
    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.
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  3. #3
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    Re: Set numbers

    Is there more of a proof way or combinatorial way to approach then just brute force counting?
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    Re: Set numbers

    Quote Originally Posted by donnagirl View Post
    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)}}
    Last edited by Plato; July 21st 2012 at 05:13 PM.
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  5. #5
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    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.
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  6. #6
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    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.
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  7. #7
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    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
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