7. $\frac{37}{56}$.

Let “ $a>b$ ” be event $A$, and “the number chosen by A contains 9” be event $B$, then $P(B)=1-P(\bar{B})=$ $1-\frac{\mathrm{C}_{8}^{3}}{\mathrm{C}_{9}^{3}}=\frac{1}{3}$.
When the number chosen by A contains 9, it is certain that $a>b$, i.e., $P(A \mid B)=1$.
When the number chosen by A does not contain 9, the sets of numbers chosen by A and B are both $\{1,2,3,4,5,6,7,8\}$, at this time the probability of $a>b$ is equal to the probability of $a<b$, and the probability of $a=b$ is $\frac{\mathrm{C}_{8}^{3}}{\left(\mathrm{C}_{8}^{3}\right)^{2}}=\frac{1}{56}$, hence $P(A \mid \bar{B})=\frac{1}{2}\left(1-\frac{1}{56}\right)=$ $\frac{55}{112}$.
Therefore, according to the total probability formula, we get
$$
P(A)=P(A \mid B) P(B)+P(A \mid \bar{B}) P(\bar{B})=1 \times \frac{1}{3}+\frac{55}{112} \times \frac{2}{3}=\frac{37}{56} .
$$
7. Person A selects three different elements from the set $\{1,2,3,4,5,6,7,8,9\}$, and arranges them in descending order to form a three-digit decimal number $a$; Person B selects three different elements from the set $\{1,2,3,4,5,6,7,8\}$, and arranges them in descending order to form a three-digit decimal number $b$. The probability that $a>b$ is $\qquad$ .