TITLE: Doubt on Interpreting Conditional Probability
QUESTION [5 upvotes]: 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
$$ P\left(R_3 ~\mid~  \left( R_1~R_2' ~\cup~ R_1'~R_2  \right) \right)
\\~\\
= \frac{P(R_3~ \cap ~\left( R_1~R_2' ~\cup~ R_1'~R_2  \right) ) }{P\left( R_1~R_2' ~\cup~ R_1'~R_2  \right)} 
\\~\\ = \frac{P(R_1 ~R_2' ~R_3 \cup  R_1 '~ R_2~ R_3)}{P\left( R_1~R_2' ~\cup~ R_1'~R_2  \right)} 
\\~\\ = \frac{P(R_1 ~R_2'~ R_3) + P(R_1 ' ~R_2~ R_3)}{P(R_1 ~R_2')  + P(R_1' ~R_2) }  
\\~\\ = \frac{\frac{3}{12}\cdot \frac{9}{11}\cdot \frac{2}{10}+  \frac{9}{12}\cdot \frac{3}{11} \cdot \frac{2}{10}}{\frac{3}{12}\cdot \frac{9}{11}+  \frac{9}{12}\cdot \frac{3}{11} } = \frac{1}{5}
$$

REPLY [2 votes]: (Expanding on Math Lover's answer, which has already identified the error.)

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.

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.