Conditional Probability Drawing Candy I figured out the first one and need help with the second, with the second one could you provide solutions for both with and without replacement?
I have two bowls of candy. One is supposed to be filled with candy-covered chocolates, the other filled with fruit-flavored candies with candy shells. Unfortunately, someone has nefariously mixed up the contents of the two bowls, mixing a portion of one bowl into the second, and vice versa. Even worse, the candies are visually indistinguishable, and can only be sorted out by taste.
I know that one of the bowls (A) has candies in the following proportions: A ~ {2/3 chocolate, 1/3 fruity}, while the other bowl (B) has candies in the following proportion: B ~ {1/4 chocolate, 3/4 fruity}. Assume for the questions below that both bowls are large enough that small samples will not have a practical effect on the distribution of candies.
I select a bowl at random and choose a random candy. It's a chocolate candy! What is the probability that I've picked up bowl A (the one weighted towards chocolates)?
ANS:24/33
Continuing with the same bowl as before, I select two more candies. They are both fruity candies. What is the probability that I've picked up bowl A (the one weighted towards chocolates) now? (To clarify, there are three candies drawn in all: one chocolate and two fruity.)

With replacement:
$$P(A|CFF) = \frac{\frac{2}{3}\cdot (\frac{1}{3})^2}{\frac{2}{3}\cdot (\frac{1}{3})^2+\frac{1}{4}\cdot (\frac{3}{4})^2} = \frac{128}{371}$$
There is an inconsistency here. That is, if both bowls are large enough not to effect the proportion ratio then solutions with and without replacement are theoretically the same. Otherwise, one would need to know the quantities in the bowls to determine the "without replacement" solution.
