You know that the final sum will be in the range of $4$ through $9$. Determine all possible rolls that yield the final result. Add up the probabilities as each event will be disjoint.
Example: to get to 4, you have the following roll patterns:
$$4 \\ 3, 1 \\ 2, 2 \\ 2, 1, 1 \\ 1, 3\\ 1,2,1 \\1, 1, 2 \\ 1, 1, 1, 1$$
This gives:
$$P(4) = P(5) = P(6) = \dfrac{1}{6}+3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{344}{1296} \\ P(7) = 3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{127}{1296} \\ P(8) = 2\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{91}{1296} \\ P(9) = \dfrac{1}{6^2}+2\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{49}{1296}$$
As a semi-verification, let's add it all up to make sure it adds to $1$:
$$\dfrac{3\cdot 343+127+91+49}{1296} = 1$$
This appears to be the correct probability for each result.
Another way to think about this: Let's consider the game in rounds. It can end on round $1$ with a roll of $4,5,6$. So that is a $\dfrac{1}{6}$ probability for each.
It can end on round $2$ with a first roll of $1,2,3$ for round $1$ to reach any of $4,5,6,7$ (so $3\dfrac{1}{6^2}$ probability of each), a first round roll of $2,3$ can reach $8$ for a probability of $2\cdot \dfrac{1}{6^2}$, or $3$ can reach $9$ for a probability of $\dfrac{1}{6^2}$.
It can end on round $3$ with the first two rolls as $(1,1),(1,2),(2,1)$ and reach $4,5,6,7,8$ for a probability of $3\cdot \dfrac{1}{6^3}$. And the first two rolls being $(1,2),(2,1)$ can reach $9$ with probability $2\cdot \dfrac{1}{6^3}$.
And it can reach any result on the fourth round by rolling three consecutive $1$'s followed by any roll of the $d6$ for a probability of $\dfrac{1}{6^4}$ each.
Best Answer
Both of them can be $5$ as well. So if the event of at least one $5$ is $B$ then,
$ \displaystyle P(B) = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac 16 \cdot \frac 16 = \frac{11}{36}$
Or using complimentary method, $ \displaystyle P(B) = 1 - \frac 56 \cdot \frac 56 = \frac{11}{36}$
Now if $A$ is the event of sum being $10$ or more,
$ \displaystyle P(A \cap B) = 2 \cdot \frac 16 \cdot \frac 16 + \frac 16 \cdot \frac 16 = \frac{3}{36}$
As there are $3$ favorable outcomes out of $36$: $\{5, 5\}, \{5, 6\}, \{6, 5\}$
So, $ \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} ~ $ does give you $\dfrac{3}{11}$