7. Randomly select three different numbers from $1,2, \cdots, 10$, the probability that they are pairwise coprime is $\qquad$
Divide $1,2, \cdots, 10$ into 6 groups: $\{1\},\{2,4,8\},\{3,9\},\{5\},\{7\},\{6,10\}$.
When selecting three numbers that are all odd, since 3 and 9 cannot be chosen simultaneously, the number of ways to choose is $C_{5}^{3}-C_{3}^{1}=7$;
Clearly, among the three selected numbers, at most one is even. When this even number is one of $2,4,8$, the two odd numbers cannot be 3 and 9 simultaneously, so the number of ways to choose is $3\left(C_{5}^{2}-1\right)=27$; When this even number is 6, the two odd numbers are chosen from $1,5,7$, so the number of ways to choose is $C_{3}^{2}=3$; When this even number is 10, the two odd numbers are chosen from $1,3,7,9$, and 3 and 9 cannot be chosen simultaneously, so the number of ways to choose is $C_{4}^{2}-1=5$.
Thus, the number of ways that satisfy the condition is $7+27+3+5=42$, let $A=$ “three different numbers are pairwise coprime”, so $P(A)=\frac{42}{C_{10}^{3}}=\frac{42}{120}=\frac{7}{20}$.