Question: I know the answer to this question is 1/10 because we know nothing about the previous two balls drawn before we draw the third ball, however Im unsure how to prove it using the law of total probability. I am able to prove it if it was the second ball drawn ie:

**P**(Draw2 is Ball3) = **P**(Draw2 is Ball3 | Draw1 is Ball3)**P**(Draw1 is Ball3) + **P**(Draw2 is Ball3|Draw1 not Ball3)**P**(Draw1 not Ball3)

The numbers end up being:

P(Draw2 is Ball3) = 0\*(1/10) + 1/9\*(9/10) = 1/10

Im not quite sure how to show it for the third draw using the same logic.

Answer: 
If we call the event "ball $i$ is $3$" $E\_i(3)$, then the probability we want is

\begin{align\*}
P(E\_3(3)) &= P(E\_3(3) \cap E\_2(3)^c \cap E\_1(3)^c) \\
 &= P(E\_3(3) | E\_2(3)^c \cap E\_1(3)^c) \\
& \times P(E\_2(3)^c | E\_1(3)^c) \\
 & \times P(E\_1(3)^c). \
\end{align\*}

From here, it's not hard to calculate the individual conditional probabilities -- for instance, $P(E\_2(3)^c | E\_1(3)^c) = 8/9$. Can you finish?
$R$ is the event that ball 3 drawn on either of the first two draws.

$S$ is the event that ball 3 drawn on the 3rd draw.

Then,

$$p(S) = [p(S|R) \times p(R)] + [p(S|\neg R) \times p(\neg R)].$$

You have already used the Law of Total Probability to compute the probability that Ball 3 was drawn on the 2nd draw. You can simply add that to the probability of Ball 3 being drawn on the 1st draw. This gives you $p(R).$