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# HELP.... Probability!

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Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7}. What is the probability that the positive difference between the two numbers is 2 or greater? Express your answer as a common fraction.

Jul 4, 2018

#1
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I will give this a try:
Since you have a set of {1, 2, 3, 4, 5, 6, 7} permuated 2 at a time, then you have a total of: 7P2 = 42 permutations as follows:
{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {1, 7} | {2, 1} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {2, 7} | {3, 1} | {3, 2} | {3, 4} | {3, 5} | {3, 6} | {3, 7} | {4, 1} | {4, 2} | {4, 3} | {4, 5} | {4, 6} | {4, 7} | {5, 1} | {5, 2} | {5, 3} | {5, 4} | {5, 6} | {5, 7} | {6, 1} | {6, 2} | {6, 3} | {6, 4} | {6, 5} | {6, 7} | {7, 1} | {7, 2} | {7, 3} | {7, 4} | {7, 5} | {7, 6} (total: 42)
I count 15 permutations of 2 numbers out of a total of 42 that the positive difference between those 15 numbers will be 2 or greater. Therefore, the probability is:
15 / 42 = 5 / 14
Note: Somebody should check this.

Jul 4, 2018
#2
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Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7}. What is the probability that the positive difference between the two numbers is 2 or greater? Express your answer as a common fraction.

If they are selected simultaneously then they could be the same number.

So there are 7*7 = 49 possible pairs.

Ho many DO NOT have a difference or 2 or more.

 1 2 3 4 5 6 7 1 0 1 2 1 0 1 3 1 0 1 4 1 0 1 5 1 0 1 6 1 0 1 7 1 0

That is 19 with a diff of 2 or less.

49-19 = 30

so the prob of the difference being 2 or greater is      $$\frac{30}{49}$$

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Jul 5, 2018
#3
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Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7}.

(7nCr2) = 21

Six pairs have an absolute difference of one (1), so (21-6) =15 have an absolute difference of 2 or greater.

15/21= 71.4%

GA

Jul 5, 2018
#4
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ok, I see your point,

The word 'simultaneously' is  not only superfluous it is also misleading.

Melody  Jul 5, 2018