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.

tanmai79
Nov 24, 2018

#1**+1 **

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

CoolStuffYT
Nov 24, 2018

#11**+1 **

Thanks, I think I made the mistake of saying you could choose 2 of the same number.

CoolStuffYT
Nov 25, 2018

#5**+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.

tanmai79
Nov 24, 2018

#8**+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

CPhill
Nov 24, 2018