Question
You have three dice which are identical except that two of them are fair and the other one is loaded, with unknown probability $p$ of coming up $6$ when rolled and there is initially equal chance that $p = \frac 1 5$ and $p = \frac 1 4$. You now randomly choose one of the three dice, roll it once, and observe that 6 comes up.
$(a)\quad$ Find the probability that the die is loaded.
$(b)\quad$ Find the probability that $p = \frac 1 4$.
My working
For $(a)$, the situation is equivalent to having three dice, $A$, $B$ and $C$, where $A$ is fair, while $B$ and $C$ have probability $\frac 1 5$ and $\frac 1 4$ of a $6$ coming up respectively and we choose $A$ with probability $\frac 2 3$ and either $B$ or $C$ with probability $\frac 1 6$ each.
Then, I drew a simple tree diagram to find that
$$\begin{aligned}
\mathbb{P}(\mathrm{Die\ is\ loaded\ } \mid 6\ \mathrm{comes\ up}) & = \frac {\left(\frac 1 6\right)\left(\frac 1 5\right) + \left(\frac 1 6\right)\left(\frac 1 4\right)} {\left(\frac 1 6\right)\left(\frac 1 5\right) + \left(\frac 1 6\right)\left(\frac 1 4\right) + \left(\frac 2 3\right)\left(\frac 1 6\right)}
\\[5 mm] & = \frac {27} {67}
\end{aligned}$$
which is the correct answer.
However, I am not sure how to approach $(b)$. The suggested solution is
$$\begin{aligned}
\mathbb{P}(p = \frac 1 4 \mid 6\ \mathrm{comes\ up}) & = \frac {\left(\frac 1 6\right)\left(\frac 1 4\right) + \left(\frac 1 2\right)\left(\frac 2 3\right)\left(\frac 1 6\right)} {\left(\frac 1 6\right)\left(\frac 1 5\right) + \left(\frac 1 6\right)\left(\frac 1 4\right) + \left(\frac 2 3\right)\left(\frac 1 6\right)}
\\[5 mm] & = \frac {35} {67}
\end{aligned}$$
I tried to use my tree diagram again to reason the working above, but I could not make sense of it. In particular, any intuitive explanations on why the solution has the $\left(\frac 1 2\right)\left(\frac 2 3\right)\left(\frac 1 6\right)$ term will be greatly appreciated 🙂
Best Answer
If, in your tree diagram, you choose the fair dice, than the unfair dice is equally likely to be unfair with $p=1/4$ or unfair with $p=1/5$. So if you got a 6 from the fair dice, you need to consider $1/2$ probability for the other unselected dice (the unfair one) to be one of the two possible $p$. Hence you get $1/2 \times \text{probability of getting a 6 from the fair dice}$
Personally, I would prefer to view the question differently than imagining three die, which seemed a bit roundabout to me. In case you think an alternative approach might be useful, let me know. Can add an addendum.
ADDENDUM
The question begins without the knowledge of what the $p$ of the unfair dice is, and says that it is equally likely to be one of the two values. Hence, the first "split" in your tree should be one of these two $p$. That is why I said that the three-dice system is a bit roundabout way to get the same "tree". If you see in the below tree, there are 4 places where we can get a 6. For the first question, you summate the probability of two of them and divide by the total probability of the 4 of them. For the second question, you summate the probability of two of them again, since now, you can select the $p=1/4$ branch and get 6 in two different ways.