Suppose that there is an urn with 3 red, 3 yellow and 3 black balls, and we are drawing without replacement.
What is the probability that the second ball is yellow, given that the first ball is either yellow or red?
I believe that this can be easily computed as:
$$\frac{1}{2}\left(\frac{2}{8}+\frac{3}{8}\right)=\frac{5}{16}$$
The reasoning is that we condition on two possible events (either yellow, or red ball drawn already), so effectively going forward, there are two sample spaces (2 yellow balls remaining and 3 yellow balls remaining): since we don't know which sample space we are in, there is $\frac{1}{2}$ chance it's either.
More formally:
- Let the event that the first ball drawn is yellow be denoted as $Y_1$
- Let the event that the first ball drawn is red be $R_1$, etc...
The probability of second yellow, given that first is yellow or red can be computed using Bayes' law as follows:
$$\mathbb{P}(Y_2|(Y_1\cup R_1))=\frac{\mathbb{P}(Y_2\cap(Y_1\cup R_1))}{\mathbb{P}(Y_1\cup R_1)}=\\=\frac{\mathbb{P}(Y_2\cap Y_1)}{\mathbb{P}(Y_1 \cup R_1)}+\frac{\mathbb{P}(Y_2\cap R_1)}{\mathbb{P}(Y_1 \cup R_1)}=\\=\frac{1}{\mathbb{P}(Y_1\cup R_1)}\left(\mathbb{P}(Y_2\cap Y_1)+\mathbb{P}(Y_2\cap R_1)\right)=\\=\frac{1}{\mathbb{P}(Y_1\cup R_1)}\left(\mathbb{P}(Y_2|Y_1)\mathbb{P}(Y_1)+\mathbb{P}(Y_2|R_1)\mathbb{P}(R_1)\right)\\=\frac{9}{6}\left(\frac{2}{8}\frac{3}{9}+\frac{3}{8}\frac{3}{9}\right)=\\=\boxed{\frac{1}{2}\left(\frac{2}{8}+\frac{3}{8}\right)}=\frac{5}{16}$$
The boxed formula shows that the "simplified" method / assumption that I show in the first half of my answer works.
The reason I am posting this question is that I have been told that my reasoning is flawed (in fact, I gave this answer to a similar question, albeit with more balls, and some commenters mentioned the logic is incorrect).
I can't see how a logic which gives the correct answer can be described as incorrect.
Your answer is correct, as confirmed by listing the restricted sample space due to the conditionality imposed with associated probabilities .
$\begin{array}{|c|c|c|c|c|c|c|}\hline Outcome & YY & RY & YR & RR & YB & RB \\ \hline Probability &\frac{6}{72} & \frac{9}{72} &\frac{9}{72}&\frac6{72} & \frac{9}{72}& \frac{9}{72}\\ \hline \end{array}$
From the table above, it should be clear that
P($2nd$ yellow |$1st$ red or yellow) $= \frac{6+9}{48} = \frac{5}{16}$