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
Answer: $\frac{9}{35}$.
Solution: A valid selection is equivalent to the total value $a$ of the two bills taken from $A$ being less than the total value $b$ of the two bills taken from $B$, thus $aa=2$, meaning the two bills taken from $B$ cannot both be 1 yuan bills, which corresponds to $\mathrm{C}_{7}^{2}-\mathrm{C}_{3}^{2}=18$ ways of selection. Therefore, the required probability is $\frac{3 \times 18}{\mathrm{C}_{5}^{2} \times \mathrm{C}_{7}^{2}}=\frac{54}{10 \times 21}=\frac{9}{35}$.