Solution verification: probability of getting tails given you got $2$ on a die in a coin-die experiment $\newcommand{\P}{\mathbb{P}}$The Problem: Consider an experiment where one first flips a fair coin. If it lands on heads, then the experimenter rolls a fair six-sided die. If it lands on tails, instead, the experimenter rolls an unfair four-sided die. (The probability of face $n$ appearing is $(5-n)/10$ on this die.)
Given that the experimenter has rolled a $2$, what is the probability that they flipped the coin and got tails just beforehand?

My Attempt: We first consider the probability of getting a $2$. This is given by
\begin{align*}
\P(\text{getting a 2})
&= \P(\text{getting heads, then a 2}) + \P(\text{getting tails, then a 2}) \\[4pt]
&= \P(\text{getting heads}) \cdot \P(\text{getting 2 on the fair die}) \\[4pt]
&\;\;\; + \P(\text{getting tails}) \cdot \P(\text{getting 2 on the unfair die}) \\[6pt]
&= \frac 1 2 \cdot \frac 1 6 + \frac 1 2 \cdot \frac{3}{10} \\
&= \frac{1}{12} + \frac{3}{20} = \frac{7}{30}
\end{align*}
Now, we note that
$$
\P(\text{getting tails} \mid \text{you got a 2}) = \frac{\P(\text{getting tails and a 2})}{\P(\text{getting a 2})}
$$
Clearly, as seen earlier, $\P(\text{getting tails and a 2}) = 3/20$. Thus,
$$
\P(\text{getting tails} \mid \text{you got a 2}) = \frac{3/20}{7/30} = \frac{9}{14} \approx 64.29\%
$$
is the probability we desire.

My Question: Basically, is this approach correct? I'm a bit rusty on my probability, so I just wanted to be sure.
My main issue for a lack of confidence is the final step here:

\begin{align*}
\P(\text{getting a 2})
&= \P(\text{getting heads, then a 2}) + \P(\text{getting tails, then a 2}) \\[4pt]
&= \P(\text{getting heads}) \cdot \P(\text{getting 2 on the fair die}) \\[4pt]
&\;\;\; + \P(\text{getting tails}) \cdot \P(\text{getting 2 on the unfair die})
\end{align*}

Is it justified to do this, i.e. to assume that the coin facing and the die value are independent and thus can be multiplied? If so, why? It feels like the two depend on each other. If not, then I'm just not sure how else to approach this.
 A: Recall that $P(A\cap B)=P(A|B)P(B)$.
Let $X$ be the flip of the coin and $Y$ be the number obtained on the second move.
Since $X=H$ and $X=T$ form a partition of the sample space, by the law of total probability $P(Y=2)=P(X=H)P(Y=2|X=H)+P(X=T)P(Y=2|X=T)$.
Define new random variables $Z$ to be the roll of the fair die, and $W$ to be the roll of the rigged die. Then $Y=Z1\{X=H\}+W1\{X=T\}$. This way, you won't get confused even if you happen to roll both die, because you know which one to choose.
Then the expression 2 paragraphs prior becomes $P(X=H)P(Z1\{X=H\}+W1\{X=T\}=2|X=H)+P(X=T)P(Z1\{X=H\}+W1\{X=T\}=2|X=T)=P(X=H)P(Z=2)+P(X=T)P(W=2)$, where I "plugged in" the value of X. Now it is clearer, if you will, that this equals $\frac 1 2 \frac 1 6+\frac 1 2\frac 3 {10}$.

Recall that the definition of independence of two events is that $P(A)=P(A|B)$, i.e. learning about the occurrence of one event has no influence on the probability of the other. Here, we found that $P(Y=2)=\frac 7 {30}$, but $P(Y=2|H=1)=\frac 1 6$. Thus the events $Y=2$ and $H=1$ are not independent. In general, $Y$ and $X$ are sample spaces and perhaps you could talk about events in the sample space as being independent.

Finally, since $X$ is defined and $Y|X$ is defined, then $X,Y$ is defined and so are $Y, X|Y$. So you could rephrase the experiment as rolling a a rigged die with probabilities $\{\frac {17}{60},\frac{7}{30},\frac{11}{60},\frac{2}{15},\frac{1}{12},\frac{1}{12}\}$, followed by rigged coins with probabilities $\{\frac{5}{17},\frac{12}{17}\},\{\frac{5}{14},\frac{9}{14}\},\{\frac{5}{11},\frac{6}{11}\},\{\frac{5}{8},\frac{3}{8}\}, \{1\}, \{1\}$, where the last two coins have heads with probability 1.
