Consider the simplified problem.
Pick an integer from 1-3 randomly. Whatever # is picked, then flip that many coins, i.e., if the integer picked, say, n=2, then we flip two coins. What is the probability that every coin flipped lands on heads?
When considering the question, I first thought of the following approach.
Idea 1: Let E be the event that all coins flipped lands on heads.
For $n=1$, $P(E)=\dfrac{1}{2}$; $n=2$, $P(E)=\dfrac{1}{2^2}$; and $n=3$, $P(E)=\dfrac{1}{2^3}$. So, my answer would be $P(E)=\dfrac{1}{2^n}$.
But a hint on the question, says, to use total probability. In that case, my answer would be
$P(E)=\sum\limits_{k=1}^3P(E|n=k)P(n=k)=\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}\right).$
My question is what in the question clearly points out which approach should I take. My thought is that the second approach needs to be taken because event $E$ results from an outcome of numerous possible events. (Second Question) If this is the correct approach, then how would the question be stated so that the first approach is correct?
The mistake on the first approach is in the notation you use. You said $$\text{Pr}(E)=\frac{1}{2^n}\ \ \text{instead of}\ \ \text{Pr}(E|n)=\frac{1}{2^n}$$ $\text{Pr}(E)$ is an unconditional probability, $\text{Pr}(E|n)$ is the conditional probability of $E$ given the number of coins tossed is $n$. And to compute $\text{Pr}(E)$ we need to consider all possibilities with an average weight given by the probabilities for each case for $n$: $$\text{Pr}(E)=\sum_{n\in \{1,2,3\}}\text{Pr}(E|n)\text{Pr}(n)$$ thats what the Law of Total Probabilities does.