<s>[INST] Probability of drawing a yellow ball after 1 or 2 have been drawn. The question is as follows:


An opaque urn contains ten balls, five red, four yellow, and one green. Three balls are      drawn from the urn in sequence but at random, without replacement.
What is the conditional probability of drawing a yellow ball given that a yellow ball has been drawn? (Note: “a yellow ball has been drawn” means at least one yellow ball has turned up, in the first draw, the second, or both.)


I have been trying to solve this problem using Bayes' Theorem the following way:
$
P(drawing\:yellow\:last\: |\: 1\: or\: 2\: drawn) = \frac{P(1\: or\: 2\: drawn\: | \:drawing \:yellow \:last)\cdot P(drawing \:yellow \:last)}{P(1 \:or \:2 \:drawn)}
$
Where $P(1 \:or \:2\: drawn\: | \:yellow\: drawn \:last)$ will be equal to $P(1\: or\: 2 \:drawn)$ because it is not a conditional probability (this may be where I'm going wrong).
$$
P(1 \:or \:2 \:drawn) = 2(\frac{4}{10}\cdot\frac{6}{9} ) + (\frac{4}{10}\cdot\frac{3}{9} ) = .666666
$$
$$
P(drawing \:yellow \:last) = 2\cdot \frac{4}{10}\cdot\frac{6}{9} \cdot \frac{3}{8} + \frac{4}{10}\cdot\frac{3}{9}\cdot \frac{2}{8} =.23333
$$
Resulting in .233333 which is not the correct answer of precisely 1/2 in the solutions given to me by my TA.
This problem is driving me insane! If you could steer me in the right direction, I'd be super grateful. Thanks!
EDIT: Here is the TA solution:

 [/INST] Let $T$ be the event yellow on third, and let $A$ be the event  at least one yellow in the first two trials. We want $\Pr(T|A)$. By the definition of conditional probability, we have
$$\Pr(T|A)=\frac{\Pr(A\cap T)}{\Pr(A)}.\tag{1}$$
We compute the two probabilities on the right-hand side of (1).
The event $A$ can happen in $3$ ways, which we can call YN, NY, and YY.
The sum of the probabilities of these is
$$\frac{4}{10}\cdot \frac{6}{9}+\frac{6}{10}\cdot\frac{4}{9}+\frac{4}{10}\cdot\frac{3}{9}.$$
The event $A\cap T$ can also happen in $3$ ways, which we can call YNY, NYY, and YYY. The probability is therefore
$$\frac{4}{10}\cdot \frac{6}{9}\cdot\frac{3}{8}+\frac{6}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}+\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8}.$$
Now we use (1) and in principle are finished, though you are probably expected to simplify. It does simplify, a lot.
</s>