6. 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, 2 bills are randomly drawn from each bag. The probability that the sum of the denominations of the remaining bills in $A$ is greater than the sum of the denominations of the remaining bills in $B$ is $\qquad$
When 2 banknotes are drawn from $A$ and their total value is less than the total value of 2 banknotes drawn from $B$, since the maximum total value of 2 banknotes drawn from $B$ is 10 yuan, the 2 banknotes drawn from $A$ cannot include a 10 yuan note. At this point, the 2 banknotes drawn from $B$ must include at least one 5 yuan note. Therefore, the probability of the event “the total value of the remaining banknotes in $A$ is greater than the total value of the remaining banknotes in $B$” is $P=\frac{C_{3}^{2}\left(C_{7}^{2}-C_{3}^{2}\right)}{C_{5}^{2} \cdot C_{7}^{2}}=\frac{3(21-3)}{10 \cdot 21}=\frac{9}{35}$.