Doubt on Interpreting Conditional Probability

I have a doubt about interpreting this question. I will post the question and my attempt. I get two different answers.

"A box contains $3$ red balls, $4$ blue balls, and $5$ white balls. Three balls are drawn at random from the box, one by one, without replacement. Find the probability that the second red ball appears on the third draw." Hence we have $3$ red and $9$ non-red balls.

If I interpret it as an unconditional probability problem, we obtain $$P(R_1 ~R_2' ~R_3 \cup R_1 '~ R_2~ R_3) \\~\\ = P(R_1 ~R_2'~ R_3) + P(R_1 ' ~R_2~ R_3) \\~\\ = \frac{3}{12}\cdot \frac{9}{11}\cdot \frac{2}{10}+ \frac{9}{12}\cdot \frac{3}{11} \cdot \frac{2}{10} = \frac{9}{110} $$ where $R_i$ is the event of drawing a red and $R_i'$ is the event of not drawing a red on the $i$'th draw.

If I interpret it as a conditional probability problem I get $$ P\left(R_3 ~\mid~ \left( R_1~R_2' ~\cup~ R_1'~R_2 \right) \right) \\~\\ = \frac{P(R_3~ \cap ~\left( R_1~R_2' ~\cup~ R_1'~R_2 \right) ) }{P\left( R_1~R_2' ~\cup~ R_1'~R_2 \right)} \\~\\ = \frac{P(R_1 ~R_2' ~R_3 \cup R_1 '~ R_2~ R_3)}{P\left( R_1~R_2' ~\cup~ R_1'~R_2 \right)} \\~\\ = \frac{P(R_1 ~R_2'~ R_3) + P(R_1 ' ~R_2~ R_3)}{P(R_1 ~R_2') + P(R_1' ~R_2) } \\~\\ = \frac{\frac{3}{12}\cdot \frac{9}{11}\cdot \frac{2}{10}+ \frac{9}{12}\cdot \frac{3}{11} \cdot \frac{2}{10}}{\frac{3}{12}\cdot \frac{9}{11}+ \frac{9}{12}\cdot \frac{3}{11} } = \frac{1}{5} $$

Of course the answers are different as both your approaches answer different questions.

In first, you are answering the probability of drawing two red balls in three draws with second red ball being fetched in the third draw. This is what the question asks.

In second, you are finding the probability of a specific outcome for only the third ball given a condition on the first two balls. The specific outcome you are seeking for the third ball in this case is red and the given condition on the first two balls is that one of them are red. You know after two draws, you are left with $10$ balls and $2$ of them are red. So the probability of third ball being red is indeed $\cfrac{2}{10} = \cfrac{1}{5}$, which is what you got applying conditional probability.

(Expanding on Math Lover's answer, which has already identified the error.)

1. In your first method, you have found this probability, as requested: $$P(RBR,RWR,BRR,WRR).$$

   Notice that identifying these $4$ favourable outcomes out of the $27$ possible ones is the simplest and most direct way to complete the given exercise, and that visualising the probability tree can be helpful.

2. In your second method, however, you are working with a reduced sample space of only $12$ outcomes, while excluding these $15$ outcomes from consideration: $$RR\_,BB\_,WW\_,BW\_,WB\_.$$ And since the favourable outcomes in this method are the same as in your first, the ensuing probability is bigger than in your first method.