There is this homework question from a yr 11 Cambridge textbook, which is:
In the textbook, the conditional probability formula for dependent events is: $P(A|B) = \frac{P(A∩B)}{P(B)}$, where $P(A|B)$ means the probability of $A$, given that $B$ has occurred, and $P(A∩B)$ means $P(A\ and\ B)$.
For the question above, I have tried using listing the possible sums, e.g $(10+5)$, $(10+10)$ and $(10+20)$ and then finding the probabilities of each without replacement, for example, to find the probability of the sum being $(10+10)$, I did $(\frac{3}{9} * \frac{2}{8})$ but I struggled to plug this into the formula.
How can I use the equation to solve the question when it is asking for the sums of the outcomes?
I'm not clear on what you tried but it sounds as if you are overcomplicating things. Sums have very little to do with it.
Let $A$ be the event that the total was \$20 or more, let $B$ be the event that at least one \$10 was picked. You want $$P(A|B)=\frac{P(A\cap B)}{P(B)}\ .$$ Now $P(B)$ should be easy enough to calculate (hint: $P(B^c)$ is probably easier), and $A\cap B$ is the event that either two \$10 or a \$10 and a \$20 are chosen.