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.
I need to find the number of ways to choose two numbers from the set such that their positive difference is at least $2$, and divide that by the total number of ways to choose two numbers from the set. The total number of ways to choose two numbers from the set is $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$. To find the number of ways to choose two numbers with a positive difference of at least $2$, I can use a systematic approach. I can start with the smallest number in the set, which is $1$, and see what other numbers I can pair it with to satisfy the condition. If I choose $1$ as the first number, the second number can be any of $3, 4, 5, 6, 7$, which gives me $5$ choices. If I choose $2$ as the first number, the second number can be any of $4, 5, 6, 7$, which gives me $4$ choices. If I choose $3$ as the first number, the second number can be any of $5, 6, 7$, which gives me $3$ choices. If I choose $4$ as the first number, the second number can be any of $6, 7$, which gives me $2$ choices. If I choose $5$ as the first number, the second number can be only $7$, which gives me $1$ choice. If I choose $6$ or $7$ as the first number, there are no more choices for the second number that satisfy the condition. Therefore, the total number of ways to choose two numbers with a positive difference of at least $2$ is $5 + 4 + 3 + 2 + 1 = 15$. The probability of choosing such a pair is $\frac{15}{21} = \frac{5}{7}$. 