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?

The probability of picking the first type of die with all purple sides equals $\Pr(T=1) = \frac{5}{20} = \frac{1}{4}$. The probability of picking the second type of die with 2 purple and 4 yellow sides equals $\Pr(T=2) = \frac{7}{20}$. The probability of picking the third type of die with 3 purple and 3 yellow sides equals $\Pr(T=3)=\frac{8}{20} = \frac{2}{5}$.
Having picked the type of die, the outcome of the second roll is independent of the first outcome. Hence:
$$
  \Pr(\mathcal{O}_2 = Y) = \Pr(T=1) \cdot \frac{0}{6} + \Pr(T=2) \frac{4}{6} + \Pr(T=3) \frac{3}{6} = \frac{13}{30} = \frac{2}{5} + \frac{1}{30} = 0.4(3)
$$
