4. Bag $A$ contains 2 ten-yuan bills and 3 one-yuan bills, and bag $B$ contains 4 five-yuan bills and 3 one-yuan bills. Now, two bills are randomly drawn from each bag. The probability that the sum of the remaining bills in $A$ is greater than the sum of the remaining bills in $B$ is $\qquad$
The number of ways to draw two banknotes from $A$ is $C_{3}^{2}+C_{2}^{1} C_{3}^{1}+C_{2}^{2}=10$, with the corresponding sum of face values being $2,11,20$; the number of ways to draw two banknotes from $B$ is $C_{3}^{2}+C_{4}^{1} C_{3}^{1}+C_{4}^{2}=21$, with the corresponding sum of face values being $2,6,10$. Thus, the condition is only met when the sum of the face values of the two banknotes drawn from $A$ is 2.

Let $C=$ “the sum of the face values of the remaining banknotes in $A$ is greater than the sum of the face values of the remaining banknotes in $B$”, so $P(C)=\frac{C_{3}^{2}\left(C_{4}^{1} C_{3}^{1}+C_{4}^{2}\right)}{10 \times 21}=\frac{3 \times 18}{10 \times 21}=\frac{9}{35}$.