Probability that the second roll comes up yellow given the first roll was purple. A bag contains $20$ dice. $5$ of the dice have entirely purple sides, $7$ of the dice have $2$ purple and $4$ yellow sides, and $8$ of the dice have $3$ purple and $3$ yellow sides. If you randomly pick a die, roll it, and observe that the roll comes up purple, what is the probability that if you roll the same die again, the roll comes up yellow?
Update I have tried the following: The probability that you pick die 1 and roll a purple is $5/20*6/6=5/20$; the probability you pick die 2 and roll a purple is $7/20*2/6=7/60$; and the probability you pick die 3 and roll a purple is $8/20*3/6=1/5$. The sum of these probabilities is $5/20+7/60+1/5=17/30$. Now the probability that the second roll is yellow given the first is purple is given by: $(5/20)\div(17/30)*0+(7/60)\div(17/30)*(4/6)+(1/5)\div(17/30)*(3/6)=.31$. This is what I think is right; can someone verify it or point out where it is wrong if it is?

Let $P$ be the event the first roll gave purple, and $Y$ the event the second roll gave yellow. We want $\Pr(Y|P)$. By the definition of conditional probability, we have
$$\Pr(Y|P)=\frac{\Pr(P\cap Y)}{\Pr(P)}.$$
You calculated $\Pr(P)$ using the correct approach. I have not checked the arithmetic. We need $\Pr(P\cap Y)$.
The event $P\cap Y$ can happen in two ways: (i) we pick a die of type 2, and roll purple then yellow or (ii) we pick a die of type 3, and roll purple then yellow. 
The probability of (i) is $\frac{7}{20}\cdot \frac{2}{6}\cdot \frac{4}{6}$. The probability of (ii) is $\frac{8}{20}\cdot \frac{3}{6}\cdot \frac{3}{6}$. Now you have all of the ingredients. 
Remark: You have obtained the same number, by essentially similar reasoning. 
