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
