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Strange Q

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

neworleans06  Nov 24, 2018
#4
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is it 5/7?

ant101  Nov 24, 2018
#6
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It is 5/7(as seen below)

tanmai79  Nov 24, 2018
#11
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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
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tanmai79  Nov 25, 2018
#5
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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
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Excellent, tanmai.....!!!!   CPhill  Nov 24, 2018
#7
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YAY! I got it! Phew, I did it! Thank you!

Nov 24, 2018
#8
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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   Nov 24, 2018
#10
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"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