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 3 or greater? Express your answer as a common fraction.
+2666 There are \({7 \choose 2} = 21 \) choices
The only cases that work are (1, 4), (1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 6), (3, 7), (4, 7)
So, the probability is \({9 \over 21} = \color{brown}\boxed{3 \over 7}\)
+2666 Yep, you're right...
I listed 10 cases but put the probability as 9/21.
