UCAT Medentry DM Question

Could someone please explain

There are 15 balls in the bag, 10 red and 5 blue. So 2/3 of the balls are red and 1/3 of the balls are blue.

If Chris takes three balls out of the bag, one at a time, without replacing them, the most likely combination of balls in his hand is 2 red balls and 1 blue ball. Are we happy with that? There are twice as many red balls in the bag as blue balls, so the most likely outcome is that Chris has twice as many red balls as blue balls. That being the case, there'd be 8 red and 4 blue balls left in the bag. So the next ball pulled is twice as likely to be red than blue.

If Chris put the balls back after he's pulled the first three balls out, then there are 10 red and 5 blue balls in the bag. So next the ball pulled is twice as likely to be red than blue.

So the different approach hasn't impacted the likelihood of the fourth ball being blue. Hence the answer is C.