Bayes' rule and probability

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I've got a probability question:

Given a 5-faced die (1,2,3,4,5),call it die $A$, each face has probability as follows:

$$\begin{array}{rrrrr} \text{Face} & 1 & 2 & 3 & 4 & 5 \\ \text{Prob} & 0.2 & 0.15 & 0.1 & 0.25 & 0.3 \end{array}$$

We roll this die three times and get $O = \{2,4,5\}$

Q1. What's the probability that we get this kind of outcome assuming that we are using die A

My solution is: $3!\cdot0.15\cdot0.25\cdot0.3$,

Q2. Given another 5-faced die $B$ and its probability distribution is as follows:

$$\begin{array}{rrrrr} \text{Face} & 1 & 2 & 3 & 4 & 5 \\ \text{Prob} & 0.1 & 0.2 & 0.3 & 0.25 & 0.15 \end{array}$$

Now, we have 2 dice, given that we do not know which die we rolled, but the outcome is $O = \{2,4,5\}$, whats the probability this die is die A?

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Probability of die A, given outcome 245, equals (probability of die A) times (probability of 245 given die A), divided by the sum of [(probability of die A) times (probability of 245 given die A)] and [(probability of die B) times (probability of 245 given die B)].

Now you have calculated probability of 245 given die A, and you can similarly calculate probability of 245 given die B, but what you need to know is the a priori probability of die A and probability of die B. Perhaps you are meant to assume that these are both 1/2.

EDIT: The above concerns Q2. I believe the solution to Q1 in the original post is correct.