7. From all four-digit numbers where each digit is unique and the sum of the digits is 10, if two numbers are randomly selected, what is the probability that 2017 is selected?
The combinations of four distinct digits that sum to 10 are $0127, 0136, 0145, 0235, 1234$, so the total number of four-digit numbers is $18 \cdot 4 + 24 = 96$. Let $A=$ “2017 is selected”, so $P(A)=\frac{95}{C_{96}^{2}}=\frac{1}{48}$.