Total sample space {b,b,b,y,y,y,} => (3blues +3 yellows) =all possible outcome
P(1st draw =b) =3/6 =1/2= 0.5 or 50%
Since it's not replaced, the sample space becomes {b,b,y,y,y}
P(2nd draw=y) = 3/5 = 0.6 = 60%
Now the probability of drawing a blue AND a yellow (blue ∩yellow), becomes a conditional probability that means P(b∩y) = p(b) x P(y)
P(blue AND yellow) = 0.5 x 0.6 = 0.3