outcome from fair and non-fair dice, dice are chosen from other dice part2 
There're 3 dice, two are fair, one is not. The non-fair dice has an occurrence of $5$ instead of $4$. That is the non-fair dice is $\{1,2,3,5,5,6\}.$
  We choose one dice from the three randomly, and throw it twice. Suppose the second roll in a throw returned $5$. What is the probability that the first roll returned $4$?

In the related question I posted earlier (outcome from fair and non-fair dice, dice are chosen from other dice part1) it was suggested that the solution will require using Bayes Rule. This is my version: $B$ - the second roll returns $5$. $A$ - the first roll returns $4$. According to Bayes rule: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{3} \times \frac{1}{6}+\frac{1}{3} \times 0}{\frac{2}{3}\times \frac{1}{6} + \frac{1}{3}\times \frac{2}{6}}$$
 A: The problem with your method is that you did not calculate $P(A \cap B)$ correctly. $P(A \cap B)$ is the probability that the first roll is $4$ and the second roll is $5$, but you just found the probability that the first roll is $4$. Let's break this up into two cases:
Case 1: Fair die


*

*The probability of getting a fair die is $\frac 2 3$

*The probability of rolling $4$ on first roll is $\frac 1 6$

*The probability of rolling $5$ on second roll is $\frac 1 6$


Case 1: Non-Fair die


*

*The probability of getting a fair die is $\frac 1 3$

*The probability of rolling $4$ on first roll is $0$

*The probability of rolling $5$ on second roll is $\frac 2 6$


Now, by the Law of Total Probability, multiply within the cases and add between the cases to get:
$$P(A \cap B)=\frac 2 3\cdot \frac 1 6\cdot \frac 1 6+\frac 1 3\cdot 0\cdot\frac 2 6=\frac 1 {54}$$
By Bayes' Rule, we get:
$$P(A | B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac 1 {54}}{\frac{2}{3}\times \frac{1}{6} + \frac{1}{3}\times \frac{2}{6}}=\frac 1 {12}$$
