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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?

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I assume the sampling of the two balls is with replacement. Because you didn't give the number of balls in each urn, numerical solutions are not possible under sampling without replacement.

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 $P(A|E)+P(B|E) = 1.$

I'm thinking you may have read this Q&A before asking your question.

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