Farina wrote:
I dont understand the solution sorry

It says 3 yellow marbles have to remain in bag, and there are already 3 yellows so we dont have to calculate or add any yellow marble? As i can see the solution for only blue marbles
We have 3 yellow and 5 blue marbles.
After 2 marbles are randomly removed, what is P(3 yellow and 3 blue marbles remain)?
In order for yellow and 3 blue marbles to REMAIN, we must REMOVE 2 blue marbles.
So, P(3 yellow and 3 blue marbles REMAIN) = P(2 blue marbles are REMOVED)
Does that help?