Question is
A box contains 2 red balls and 4 yellow balls. If a ball is randomly drawn from the box, find the probability that it is (a)(i) a red ball, (a)(ii) yellow ball.

I can handle (a)(i) and (ii), obvious answers 2/6 and 4/6.

But I get stuck at part (b) which reads:

If 2 balls are randomly chosen and simultaneously removed from the box, what is the probability that only yellow balls are left in the box?

Please correct me if I am wrong,
The total possible outcomes for the 2 balls removed from the box are
{1 red 1 yellow,
2 red 0 yellow,
0 red 2 yellow}

At this point, I am not sure if I am right, if only yellow balls are left in the box that means all red ball (i.e. 2 red balls) are removed, and according to the above total possible outcomes, the probability is 1/3, am I correct?

The way you listed the events are not wrong, but you need to know that they are not equally-likely to happen - so you cannot jump to the conclusion that the probability is one-third.

If you have learned some combinatorics, you should know how to calculate the number of ways to pick 2 balls from a total of 6 balls. And that should help you to get the answer.

After you are able to handle this part, you should go to learn the hypergeometric distribution which help you to solve all these kind of elementary problems.

So, you're saying if the event "2 balls are randomly and simultaneously removed from the box" is equally likely to happen, then my calculation is correct?