smc

This is not too bad using casework. Tina gets a sum of 3: This happens in only one way $(1,2)$ and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here. Tina gets a sum of 4: This once again happens in only one way $(1,3)$. Sergio can choose a number from 5 to 10, so 6 ways here. Tina gets a sum of 5: This can happen in two ways $(1,4)$ and $(2,3)$. Sergio can choose a number from 6 to 10, so $2\cdot 5=10$ ways here. Tina gets a sum of 6: Two ways here $(1,5)$ and $(2,4)$. Sergio can choose a number from 7 to 10, so $2\cdot 4=8$ here. Tina gets a sum of 7: Two ways here $(2,5)$ and $(3,4)$. Sergio can choose from 8 to 10, so $2\cdot 3=6$ ways here. Tina gets a sum of 8: Only one way possible $(3,5$). Sergio chooses 9 or 10, so 2 ways here. Tina gets a sum of 9: Only one way $(4,5)$. Sergio must choose 10, so 1 way. In all, there are $7+6+10+8+6+2+1=40$ ways. Tina chooses two distinct numbers in $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ ways while Sergio chooses a number in $10$ ways, so there are $10\cdot 10=100$ ways in all. Since $\frac{40}{100}=\frac{2}{5}$, our answer is $\boxed{2/5}$.