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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 $$ P\left(R_3 ~\mid~ \left( R_1~R_2' ~\cup~ R_1'~R_2 \right) \right) \\~\\ = \frac{P(R_3~ \cap ~\left( R_1~R_2' ~\cup~ R_1'~R_2 \right) ) }{P\left( R_1~R_2' ~\cup~ R_1'~R_2 \right)} \\~\\ = \frac{P(R_1 ~R_2' ~R_3 \cup R_1 '~ R_2~ R_3)}{P\left( R_1~R_2' ~\cup~ R_1'~R_2 \right)} \\~\\ = \frac{P(R_1 ~R_2'~ R_3) + P(R_1 ' ~R_2~ R_3)}{P(R_1 ~R_2') + P(R_1' ~R_2) } \\~\\ = \frac{\frac{3}{12}\cdot \frac{9}{11}\cdot \frac{2}{10}+ \frac{9}{12}\cdot \frac{3}{11} \cdot \frac{2}{10}}{\frac{3}{12}\cdot \frac{9}{11}+ \frac{9}{12}\cdot \frac{3}{11} } = \frac{1}{5} $$

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2 Answers 2

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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.

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    $\begingroup$ I want to disagree with your answer (respectfully), and ask why it will not be $\frac{9}{110}$. If one looks through a tree diagram, there are $2$ possible branches the condition holds true ($R-R'-R$ or $R'-R-R$), therefore you can add their probabilities up. Also, I did not quite get the understanding of "In first, you are answering the probability of drawing second red ball in third draw" ; I believe we are already accounting for $1$ red ball already being drawn through the $2$ possible branches - (or am I getting this wrong?) $\endgroup$ Sep 12, 2021 at 4:30
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    $\begingroup$ @SnipingPoodle for the given question, $\frac{9}{110}$ is the correct answer. Sorry if it did not come across that way in my explanation. I will edit. $\endgroup$
    – Math Lover
    Sep 12, 2021 at 4:37
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    $\begingroup$ because $P\left(R_3 ~\cap \left( R_1~R_2' ~R_3 ~\cup~ R_1'~R_2 ~R_3 \right) \right)$ is $P\left( R_1~R_2' ~R_3 ~\cup~ R_1'~R_2 ~R_3 \right)$ $\endgroup$
    – Math Lover
    Sep 12, 2021 at 5:00
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    $\begingroup$ @MathLover I see since $R_3 \cap R_3 = R_3$ so then this is not a 'conditional' probability question per se, and my attempt to cast it as such didn't work. Thus the unconditional probability interpretation is correct. I had a quick question, do you agree that $P\left(R_3 ~\mid~ \left( R_1~R_2' ~\cup~ R_1'~R_2 \right) \right)$ is equivalent to $P\left(R_3 ~\mid~ \left( R_1~R_2' ~R_3~\cup~ R_1~R_2' ~R_3' \cup R_1'~R_2 ~R_3 \cup R_1'~R_2 ~R_3' \right) \right)$. This helps me think what $P (R_3 ~\mid~ ( R_1~R_2' ~\cup~ R_1'~R_2 )) $ actually means. After expanding some terms go to zero. $\endgroup$
    – john
    Sep 12, 2021 at 5:04
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    $\begingroup$ Yes your interpretation of $P\left(R_3 ~\mid~ \left( R_1~R_2' ~\cup~ R_1'~R_2 \right) \right)$ is correct. Also yes, it is not a conditional probability question. $\endgroup$
    – Math Lover
    Sep 12, 2021 at 5:10
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(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.

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