# Probability of selecting from a bin

if there are 4 bins labelled A and 6 bins labelled B each with two balls: distribution of balls in A (0.1, 0.3, 0.2, 0.4) Red, Blue, White Black distribution of balls in B (0.4, 0.2, 0.3, 0.1) Red, Blue, White Black and again you pick from a bin and get one red and blue, the probability it is from a bin labelled A has to be much lower than 0.4 right?

Let $$E$$ be the event that in two draws for an urn you get exactly one red and one blue ball. Then $$P(E|A) = 1(.1)(.3) = 0.06$$ and $$P(E|B) = 2(.4)(.2) = 0.16.$$ Also, I assume the choice of urn is random so that $$P(A) = 0.4$$ and $$P(B) = 0.6.$$
You seek $$P(A|E) = P(AE)/P(E) = \frac{P(A)P(E|A)}{P(A)P(E|A)+P(B)P(E|B)}\\ = \frac{.4(.06)}{.4(.06)+.6(.16)} = \frac{.024}{.024+.096} = 0.2$$ and $$P(B|E) = P(BE)/P(E) = \frac{P(B)P(E|B)}{P(A)P(E|A)+P(B)P(E|B)}\\ = \frac{.6(.16)}{.4(.06)+.6(.16)} = \frac{.096}{.024+.096} = 0.8.$$
So you're correct that the first conditional probability is smaller. Actually, it is not necessary to compute the second probability because $$P(A|E)+P(B|E) = 1.$$