7. It is known that 60 students from schools $A, B, C$ participate in an interview, with 10 students from school $A$, 20 students from school $B$, and 30 students from school $C$. During the interview, a student is randomly selected from those who have not yet been interviewed, and after the interview, the next student is chosen. Then the probability that all students from school $A$ complete their interviews before students from the other two schools is $\qquad$ .
When the last interviewee is a student from school $B$, removing all students from school $B$, the student at the last position must be from school $C$, the probability of this event is $p_{1}=\frac{20}{60} \times \frac{30}{40}=\frac{1}{4}$;
When the last interviewee is a student from school $C$, removing all students from school $C$, the student at the last position must be from school $B$, the probability of this event is $p_{2}=\frac{30}{60} \times \frac{20}{30}=\frac{1}{3}$.
Therefore, the probability that students from school $A$ complete their interviews before students from the other two schools is $p=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$.