You're entirely correct. Given that you've drawn two white balls, you've got the following configuration in your bag: 6 white balls, 5 non-white balls. The probability of drawing a third white ball is $\frac{6}{11}$.
"At first glance you may say 1/3, as you hold a yellow ball so 1/3 chance one of the remaining three balls will be yellow."
Well, that assumes you have a specific ball in mind to be "the" yellow that is yellow and a specific ball in mind to be the "other" ball. But you don't. Either ball (or both) can be "the" yellow ball and either (or both) can be the "other".
Think of it this way: If you say the ball in his right hand is yellow then the probability that the ball in his left hand is yellow is 1/3. And if you say the ball in his left hand is yellow then the probability that the ball in right hand is yellow is 1/3. But you are asking if either the ball in his left hand or the ball in his right is yellow what is the probability that the balls in both hands are yellow.
Those are obviously different questions.
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Actually your list should have 12, not 7, options.
$Y_1Y_2, Y_1B, Y_1R, Y_2Y_1, Y_2B, Y_2R, BY_1, BY_2, BR, RY_1,RY_2, RB$.
And of the $10$ remaining there are $2$ options.
Best Answer
I have solved it in two ways. First I solved it for all places where blue ball start drawing. And I get,
$= \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84}$
$= \frac{7}{84} = \frac{1}{12}$
Second method,
Group 3 blue balls as 1 ball. So now total balls 7.
So favourable cases 7! * 3!.
3! for blue balls.
Total cases 9!.
Probability = $\frac{7! * 3!}{9!}$
=$ \frac{1}{12} = .083$
You can apply this method to check other example.