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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.

 Nov 24, 2018
 #1
avatar+1035 
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There are \(7C2=\frac{7!}{5!2!}=21\)  possibilities of number pairs.

For the difference to be 2 or greater, the numbers cannot be 1 apart. Therefore, the sets {1, 2}, {2, 3}, {3, 4}, etc. all the way to {6, 7} cannot be used. Counting, we have 6 pairs that don't work so \(21-6=15\) and the probability is \(\frac{15}{21}=\frac{5}{7}\).

 

You are very welcome!

:P

 Nov 24, 2018
edited by CoolStuffYT  Nov 25, 2018
 #2
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+1

This answer is incorrect.

Guest Nov 24, 2018
 #3
avatar+52 
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Try again; you're close through. 

neworleans06  Nov 24, 2018
 #4
avatar+878 
+2

is it 5/7?

ant101  Nov 24, 2018
 #6
avatar+142 
+1

It is 5/7(as seen below)

tanmai79  Nov 24, 2018
 #11
avatar+1035 
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Thanks, I think I made the mistake of saying you could choose 2 of the same number.

CoolStuffYT  Nov 25, 2018
 #12
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Now the answer isn't incorrect.

tanmai79  Nov 25, 2018
 #5
avatar+142 
+2

I found it:

 

The only time the two numbers selected will not have a positive difference which is 2 or greater is when the two numbers are consecutive. There are 6 pairs of consecutive numbers in the set, and there are (\($\binom{7}{2}=21$\) pairs of numbers total. So, the probability that the pair of numbers chosen is not consecutive is  1-6/21=5/7.

 Nov 24, 2018
 #9
avatar+107405 
+1

Excellent, tanmai.....!!!!

 

 

cool cool cool

CPhill  Nov 24, 2018
 #7
avatar+878 
+3

YAY! I got it! Phew, I did it! Thank you!

 Nov 24, 2018
 #8
avatar+107405 
+2

We have   C(7,2)  = 21 possible sets formed by choosing any two of the seven numbers

 

There are 6 sets  where the positive difference is < 2

(1,2) (2,3) (3,4) (4,5) (5,6) (6, 7)

 

So....the probability that two numbers are drawn whose difference is ≥ 2 is

 

[ 21 - 6 ] / 21    =   15/ 21   =   5 / 7

 

 

cool cool cool

 Nov 24, 2018
 #10
avatar+142 
+1

"Yays" all around!

 

( I wanted to put a picture of an emoji with a body putting his hands in the air here, but it's not letting me)

 Nov 24, 2018

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