Probably that an $80\%$-truthful person actually rolled a $6$ A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?
I know there is a similar question like this 
but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also.
In my attempt, I defined the events
\begin{align*}
 E_1&: \text{The person tells the truth.} \\
 E_2&: \text{The person lies.} \\
 E_3&: \text{The person reports that the die landed on a 6.} 
\end{align*}
I noted that $P(E_1)=\frac{4}{5}$, $P(E_2)=\frac{1}{5}$, $P(E_3|E_1)=6^{-1}$ and
$P(E_3|E_2)=0$ and obtained
\begin{align*}
 P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}.
\end{align*}
However, the correct answer is, $\frac{4}{9}$. What did I do wrong?
Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events.

Let $D_{6}$ denote the event that the die lands on a $6$ and $R_{6}$ denote the event that the person reports that it landed on a $6$. You know that
\begin{align*}
 P(R_{6}|D_{6})     &= 0.8 \\
 P(D_{6})           &= 6^{-1}
\end{align*}
In order to obtain the correct answer, you also need to assume that $P(R_{6}|D_{6}^{c})=0.2$, that is, if the person lies and doesn't obtain a $6$, then he says he obtained a $6$.
Since you wish to determine $P(D_{6}|R_{6})$, which is an inverse probability of $P(R_{6}|D_{6})$, a strategy that usually works is applying Bayes' theorem.
\begin{align*}
 P(D_{6}|R_{6}) 
 &= \frac{P(D_{6})P(R_{6}|D_{6})}
{P(D_{6})P(R_{6}|D_{6})+P(D_{6}^c)P(R_{6}|D_{6}^c)} \\
 &= \frac{6^{-1} \cdot 0.8}{6^{-1} \cdot 0.8 + 5 \cdot 6^{-1} \cdot 0.2}
 = \frac{4}{9}
\end{align*}
